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Search: id:A064943
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| A064943 |
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Number of integers with 2*n digits that are the sum of the squares of their halves (leading zeros count, 1 does not to avoid the ambiguity 1=0^2+1^2 = 00^2+01^2 = 000^2+001^2 = ...). |
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+0
2
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| 0, 2, 2, 2, 6, 6, 14, 30, 6, 14, 14, 6, 6, 14, 126, 14, 14, 62, 6, 14, 126, 14, 14, 510, 126, 14, 62, 30, 30, 62, 6, 6, 254, 14, 2046, 30, 126, 62, 126, 510, 6, 254, 6, 14, 2046, 14, 14, 254, 30, 254, 2046, 254, 30, 254, 4094, 510, 2046, 126, 6, 254, 30, 126, 2046, 14
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Is there any n > 1 with a(n) = 0 ? This is equivalent to the question whether there is any prime of the form 10^(2*n)+1 besides 10^(2*1)+1 = 101. If such a prime exists, n must be a power of 2. Up to now no such prime is known.
68 is the smallest n where a(n) is not a power of two minus 2 (a(68)=22) since (10^136)+1 is the smallest integer among the 10^(2*n)+1 which is not square-free (10^136+1 = 17^2 * P7 * P11 * P117, so tau(10^136+1) = 24)
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LINKS
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Cunningham project factorization tables of 10^k+1
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FORMULA
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tau (10^(2*n)+1)-2
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EXAMPLE
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a(5) = 6 because of 1765038125 = 17650^2+38125^2, 2584043776 = 25840^2+43776^2, 7416043776 = 74160^2+43776^2, 8235038125 = 82350^2+38125^2, 9901009901 = 99010^2+09901^2, 99009901 = 00990^2+09901^2 (the last one counts as 10-digit number) Alternatively: a(5) = tau(10^(2*5)+1)-2 = tau(101*3541*27961)-2 = 8-2 = 6
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CROSSREFS
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Cf. A064942 and A002654 for the derivation of the formula
Sequence in context: A155453 A032558 A139552 this_sequence A081478 A105341 A151694
Adjacent sequences: A064940 A064941 A064942 this_sequence A064944 A064945 A064946
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KEYWORD
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nonn,base
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AUTHOR
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Ulrich Schimke (ulrschimke(AT)aol.com)
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