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Search: id:A068009
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| A068009 |
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Square array T(m,n) with m (row) >= 1 and n (column) >= 0 read by antidiagonals: number of subsets of {1,2,3,...n} that sum to 0 mod m. (Including the empty set whose sum is 0). |
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25
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| 1, 2, 1, 4, 1, 1, 8, 2, 1, 1, 16, 4, 2, 1, 1, 32, 8, 4, 1, 1, 1, 64, 16, 6, 2, 1, 1, 1, 128, 32, 12, 4, 2, 1, 1, 1, 256, 64, 24, 8, 4, 2, 1, 1, 1, 512, 128, 44, 16, 8, 3, 1, 1, 1, 1, 1024, 256, 88, 32, 14, 6, 3, 1, 1, 1, 1, 2048, 512, 176, 64, 26, 12, 5, 2, 1, 1, 1, 1, 4096, 1024, 344, 128
(list; table; graph; listen)
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OFFSET
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0,2
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COMMENT
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When p is an odd prime, T(p,k+p) = 2*T(p,k) + (2^k * ((2^p) - 2)/p) for all k >= 0 [Sophie LeBlanc]
When m divides n (with n >= m), T(m,n) = (1/m) Sum_{d | m and d is odd} phi(d) * 2^(n/d) [N. Kitchloo and L. Pachter; D. Rusin]
A068009[C(i+1,2),i] = 2, A068009[C(i,2)+1,i] = A000009[i-1]+1 [AK, cf. A068049]
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REFERENCES
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N. Kitchloo and L. Pachter, An interesting result about subset sums.
Bill Pet, Sophie LeBlanc, Will Self et al., 2002 [See the sci.math thread given above]
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LINKS
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A. Karttunen, Scheme code for computing this table and its rows.
Lior Pachter, Subset sums
Bill Pet, Sophie LeBlanc, Will Self et al., Subsets of {1,2,3,...,n} (discussion in sci.math)
Index entries for sequences related to subset sums mod m
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CROSSREFS
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Main diagonal: A000016, super-diagonal: A063776. The first term greater than one occurs on each row m in the position A002024[m] and these are given in A068049.
Row 1: A000079, row 2: A011782, row 3: A068010, row 5: A068011, row 6: A068012, row 7: A068013, row 9: A068030, row 10: A068031, row 11: A068032, row 12: A068033, row 13: A068034, row 14: A068035, row 15: A068036, row 16: A068037, row 17: A068038, row 18: A068039, row 19: A068040, row 20: A068041, row 21: A068042, row 25: A068043, row 32: A068044, row 64: A068045.
Sequence in context: A088443 A117352 A137710 this_sequence A140168 A059119 A127772
Adjacent sequences: A068006 A068007 A068008 this_sequence A068010 A068011 A068012
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KEYWORD
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nonn,nice,tabl
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AUTHOR
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This entry and Scheme-code created by Antti Karttunen, Feb 11 2002
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