A141534 Derived from the centered polygonal numbers: start with the first triangular number, then the sum of the first square number and the second triangular number, then the sum of first pentagonal number, the second square number and the third triangular number, and so on and so on...

1, 4, 11, 26, 55, 105, 184, 301, 466, 690, 985, 1364, 1841, 2431, 3150, 4015, 5044, 6256, 7671, 9310, 11195, 13349, 15796, 18561, 21670, 25150, 29029, 33336, 38101, 43355, 49130, 55459, 62376, 69916, 78115, 87010, 96639, 107041, 118256, 130325

Offset

1,2

Comments

Consider the array of triangular, square and centered polygonal numbers (irregular variant of A086272 and A086273):

1 3 6 10 15 21 28 36 45 55 A000217

1 4 9 16 25 36 49 64 81 100 A000290

1 6 16 31 51 76 106 141 181 226 A005891

1 7 19 37 61 91 127 169 217 271 A003215

1 8 22 43 71 106 148 197 253 316 A069099

1 9 25 49 81 121 169 225 289 361 A016754

1 10 28 55 91 136 190 253 325 406 A060544

1 11 31 61 101 151 211 281 361 451 A062786

1 12 34 67 111 166 232 309 397 496 A069125

1 13 37 73 121 181 253 337 433 541 A003154

1 14 40 79 131 196 274 365 469 586 A069126

1 15 43 85 141 211 295 393 505 631 A069127

etc. The sequence contains the antidiagonal sums of this array. - R. J. Mathar, Jun 05 2011

For comparison, the antidiagonal sums of A086270 are essentially A006522 starting at the 4th term. - R. J. Mathar, Sep 20 2008

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = (n-1)*(n^3+11*n^2-38*n+120)/24, n>1. - R. J. Mathar, Sep 12 2008

G.f.: x*(1-x+x^2+x^3-x^5)/(1-x)^5. - Alexander R. Povolotsky, Jun 06 2011

Crossrefs

Cf. A000217.

Sequence in context: A076048 A109414 A027966 * A320852 A192961 A290989

Adjacent sequences: A141531 A141532 A141533 * A141535 A141536 A141537

Keyword

nonn,easy

Author

Dan Graybill (clopen(AT)comcast.net), Aug 12 2008

Status

approved