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 A215573 a(n) = n*(n+1)*(2n+1)/6 modulo n. 4
 0, 1, 2, 2, 0, 1, 0, 4, 6, 5, 0, 2, 0, 7, 10, 8, 0, 3, 0, 10, 14, 11, 0, 4, 0, 13, 18, 14, 0, 5, 0, 16, 22, 17, 0, 6, 0, 19, 26, 20, 0, 7, 0, 22, 30, 23, 0, 8, 0, 25, 34, 26, 0, 9, 0, 28, 38, 29, 0, 10, 0, 31, 42, 32, 0, 11, 0, 34, 46, 35, 0, 12, 0, 37, 50, 38 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS a(n) = 0 for n = 6k +- 1, that is, A007310 (numbers congruent to 1 or 5 mod 6). Graph consists of 4 linear patterns. LINKS Zak Seidov, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1). FORMULA a(n) = A000330(n) mod n. From Colin Barker, Feb 07 2019: (Start) G.f.: x^2*(1 + 2*x + 2*x^2 + x^4 + 2*x^6 + 2*x^7 + x^8) / ((1 - x)^2*(1 + x)^2*(1 - x + x^2)^2*(1 + x + x^2)^2). a(n) = 2*a(n-6) - a(n-12) for n>12. (End) a(6*n) = n, a(6*n+1) = 0, a(6*n+2) = 3*n+1, a(6*n+3) = 4*n+2, a(6*n+4) = 3*n+2, a(6*n+5) = 0. - Philippe Deléham, Mar 05 2023 a(n) = A048153(n) mod n. - Alois P. Heinz, Jun 03 2024 a(n) = A000330(n-1) mod n. - Chai Wah Wu, Jun 03 2024 MAPLE seq(modp(n*(n+1)*(2*n+1)/6, n), n=1..100); # Muniru A Asiru, Feb 07 2019 MATHEMATICA Table[Mod[(n(n+1)(2n+1))/6, n], {n, 80}] (* or *) LinearRecurrence[{0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, -1}, {0, 1, 2, 2, 0, 1, 0, 4, 6, 5, 0, 2}, 80] (* Harvey P. Dale, Aug 25 2023 *) PROG (PARI) a(n)=n*(n+1)*(2*n+1)/6 % n; \\ Michel Marcus, Oct 19 2013 (PARI) concat(0, Vec(x^2*(1 + 2*x + 2*x^2 + x^4 + 2*x^6 + 2*x^7 + x^8) / ((1 - x)^2*(1 + x)^2*(1 - x + x^2)^2*(1 + x + x^2)^2) + O(x^80))) \\ Colin Barker, Feb 07 2019 (Python) def A215573(n): return n*(n-1)*((n<<1)-1)//6%n # Chai Wah Wu, Jun 03 2024 CROSSREFS Cf. A000330, A007310, A048153. Sequence in context: A251690 A187752 A295181 * A163537 A219946 A117449 Adjacent sequences: A215570 A215571 A215572 * A215574 A215575 A215576 KEYWORD nonn,easy AUTHOR Zak Seidov, Aug 16 2012 STATUS approved

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Last modified July 24 02:45 EDT 2024. Contains 374575 sequences. (Running on oeis4.)