A093722 Integers of the form (k^2 - 1) / 120.
0, 1, 3, 7, 8, 14, 20, 29, 31, 42, 52, 66, 69, 85, 99, 118, 122, 143, 161, 185, 190, 216, 238, 267, 273, 304, 330, 364, 371, 407, 437, 476, 484, 525, 559, 603, 612, 658, 696, 745, 755, 806, 848, 902, 913, 969, 1015, 1074, 1086, 1147, 1197, 1261, 1274, 1340
Offset: 1
Links
- Harry J. Smith, Table of n, a(n) for n = 1..20000
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,2,-2,0,0,-1,1).
Programs
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Maple
A093722 := proc(q) local n; for n from 0 to q do if type(sqrt(120*n+1), integer) then print(n); fi; od; end: A093722(1500); # Peter Bala, Dec 26 2024
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Mathematica
Select[Table[(n^2-1)/120,{n,0,700}],IntegerQ] (* Harvey P. Dale, Nov 26 2010 *)
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PARI
{a(n) = (((n\4 * 3 + n%4) * 10 + (-1)^(n\2))^2 - 1) / 120 } /* Michael Somos, Oct 17 2006 */
Formula
a(-1 - n) = a(n). a(n) = (A057538(n) * 2 - 1) / 120.
G.f.: -x^2*(1+2*x+4*x^2+x^3+4*x^4+x^6+2*x^5) / ( (1+x)^2*(x^2+1)^2*(x-1)^3 ). - R. J. Mathar, Jun 09 2013
From Peter Bala, Dec 26 2024: (Start)
a(n) is quasi-polynomial in n
a(4*n) = n*(15*n + 1)/2; a(4*n+1) = (3*n + 1)*(5*n + 2)/2;
a(4*n+2) = (3*n + 2)*(5*n + 3)/2; a(4*n+3) = (n + 1)*(15*n + 14)/2.
For 0 <= k <= 3, a(4*n+k) = (N_k(n)^2 - 1)/120, where N_0(n) = 30*n + 1, N_1(n) = 30*n + 11, N_2(n) = 30*n + 19 and N_3(n) = 30*n + 29. (End)
Extensions
More terms from Harvey P. Dale, Nov 26 2010
Offset corrected to 1 by Ray Chandler, Jul 29 2019
Comments