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Search: a263615 -id:a263615
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a(2n) = A000125(n), a(2n+1) = 2*a(2n).
+0
2
0, 0, 1, 2, 2, 4, 4, 8, 8, 16, 15, 30, 26, 52, 42, 84, 64, 128, 93, 186, 130, 260, 176, 352, 232, 464, 299, 598, 378, 756, 470, 940, 576, 1152, 697, 1394, 834, 1668, 988, 1976, 1160, 2320, 1351, 2702, 1562, 3124, 1794, 3588, 2048, 4096, 2325, 4650, 2626, 5252, 2952, 5904, 3304, 6608, 3683, 7366
OFFSET
0,4
COMMENTS
For n >= 2, number of palindromic squares of length n whose decimal digits are 0 or 1 and with 9 or fewer 1's.
LINKS
G. J. Simmons, Palindromic powers, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy]
FORMULA
From Colin Barker, Oct 26 2015: (Start)
a(n) = (-((-1)^n*(-78+62*n-12*n^2+n^3))+3*(-26+42*n-8*n^2+n^3))/96.
a(n) = 4*a(n-2)-6*a(n-4)+4*a(n-6)-a(n-8) for n>7.
G.f.: x^2*(2*x+1)*(2*x^4-2*x^2+1) / ((x-1)^4*(x+1)^4).
(End)
PROG
(PARI) a(n) = (-((-1)^n*(-78+62*n-12*n^2+n^3))+3*(-26+42*n-8*n^2+n^3))/96 \\ Colin Barker, Oct 26 2015
(PARI) concat(vector(2), Vec(x^2*(2*x+1)*(2*x^4-2*x^2+1)/((x-1)^4*(x+1)^4) + O(x^100))) \\ Colin Barker, Oct 26 2015
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
N. J. A. Sloane, Oct 23 2015
STATUS
approved

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