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 A256645 25-gonal pyramidal numbers: a(n) = n*(n+1)*(23*n-20)/6. 9
 0, 1, 26, 98, 240, 475, 826, 1316, 1968, 2805, 3850, 5126, 6656, 8463, 10570, 13000, 15776, 18921, 22458, 26410, 30800, 35651, 40986, 46828, 53200, 60125, 67626, 75726, 84448, 93815, 103850, 114576, 126016, 138193, 151130, 164850, 179376, 194731, 210938 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS If b(n,k) = n*(n+1)*((k-2)*n-(k-5))/6 is n-th k-gonal pyramidal number, then b(n,k) = A000292(n) + (k-3)*A000292(n-1) (see Deza in References section, p. 96). Also, b(n,k) = b(n,k-1) + A000292(n-1) (see Deza in References section, p. 95). Some examples: for k=4, A000330(n) = A000292(n) + A000292(n-1); for k=5, A002411(n) = A000330(n) + A000292(n-1); for k=6, A002412(n) = A002411(n) + A000292(n-1), etc. This is the case k=25. REFERENCES E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93 (23rd row of the table). LINKS Colin Barker, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1). FORMULA G.f.: x*(1 + 22*x)/(1 - x)^4. a(n) = A000292(n) + 22*A000292(n-1) = A256716(n) + A000292(n-1), see comments. a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3. - Colin Barker, Apr 07 2015 MATHEMATICA Table[n (n + 1) (23 n - 20)/6, {n, 0, 40}] LinearRecurrence[{4, -6, 4, -1}, {0, 1, 26, 98}, 40] (* Vincenzo Librandi, Apr 08 2015 *) PROG (PARI) concat(0, Vec(x*(1 + 22*x)/(1 - x)^4 + O(x^100))) \\ Colin Barker, Apr 07 2015 (MAGMA) k:=25; [n*(n+1)*((k-2)*n-(k-5))/6: n in [0..40]]; // Vincenzo Librandi, Apr 08 2015 CROSSREFS Partial sums of A255184. Cf. similar sequences listed in A237616. Cf. A000292, A256716. Sequence in context: A038654 A010014 A095796 * A175549 A159541 A144129 Adjacent sequences:  A256642 A256643 A256644 * A256646 A256647 A256648 KEYWORD nonn,easy AUTHOR Luciano Ancora, Apr 07 2015 STATUS approved

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Last modified February 20 15:10 EST 2019. Contains 320337 sequences. (Running on oeis4.)