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Search: id:A112574
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| A112574 |
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G.f. A(x) satisfies: A(x)^2 equals the g.f. of A110649, which consists entirely of numbers 1 through 12. |
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1
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| 1, 3, 0, 2, 0, 3, -8, 30, -90, 290, -930, 3000, -9696, 31461, -102420, 334467, -1095510, 3598464, -11852026, 39136629, -129548493, 429817733, -1429178703, 4761992751, -15898024868, 53174651133, -178168302693, 597971203902, -2010093276240, 6767100270918
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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A110649 is formed from every 2-nd term of A084067, which also consists entirely of numbers 1 through 12.
Why are so many a(n) divisible by 3, i.e. 22 of the first 28? - Jonathan Vos Post (jvospost2(AT)yahoo.com), Sep 14 2005
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FORMULA
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G.f. A(x) satisfies: A(x)^4 (mod 8) = g.f. of A084067.
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EXAMPLE
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A(x) = 1 + 3*x + 2*x^3 + 3*x^5 - 8*x^6 + 30*x^7 - 90*x^8 +..
A(x)^2 = 1 + 6*x + 9*x^2 + 4*x^3 + 12*x^4 + 6*x^5 +...
A(x)^4 = 1 + 12*x + 54*x^2 + 116*x^3 + 153*x^4 + 228*x^5 +..
A(x)^4 (mod 8) = 1 + 4*x + 6*x^2 + 4*x^3 + x^4 + 4*x^5 +...
G(x) = 1 + 12*x + 6*x^2 + 4*x^3 + 9*x^4 + 12*x^5 + 4*x^6 +..
where G(x) is the g.f. of A084067.
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PROGRAM
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(PARI) {a(n)=local(d=2, m=12, A=1+m*x); for(j=2, d*n, for(k=1, m, t=polcoeff((A+k*x^j+x*O(x^j))^(1/m), j); if(denominator(t)==1, A=A+k*x^j; break))); polcoeff(Ser(vector(n+1, i, polcoeff(A, d*(i-1))))^(1/2), n)}
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CROSSREFS
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Cf. A110649, A084067.
Adjacent sequences: A112571 A112572 A112573 this_sequence A112575 A112576 A112577
Sequence in context: A092735 A035464 A019746 this_sequence A061064 A016597 A128114
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KEYWORD
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sign
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Sep 14 2005
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