cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A093918 a(2k-1)=(2k-1)^2+k, a(2k)=6k^2+k+1: Last term in rows of triangle A093915.

Original entry on oeis.org

2, 8, 11, 27, 28, 58, 53, 101, 86, 156, 127, 223, 176, 302, 233, 393, 298, 496, 371, 611, 452, 738, 541, 877, 638, 1028, 743, 1191, 856, 1366, 977, 1553, 1106, 1752, 1243, 1963, 1388, 2186, 1541, 2421, 1702, 2668, 1871, 2927, 2048, 3198, 2233, 3481, 2426, 3776
Offset: 1

Views

Author

Amarnath Murthy, Apr 25 2004

Keywords

Comments

Initially defined as "leading diagonal" of the triangle A093915, a(n) is the last term in row n of A093915, i.e. a(n)=A093916(n)+n-1. - M. F. Hasler, Apr 04 2009

Links

Crossrefs

Cf. A093915, A093916, A093917.

Programs

  • Mathematica
    LinearRecurrence[{0,3,0,-3,0,1},{2,8,11,27,28,58},50] (* Harvey P. Dale, Oct 22 2013 *)
  • PARI
    A093918(n)=if(n%2,n^2,6*(n\2)^2)+n\2+1 \\ M. F. Hasler, Apr 04 2009

Formula

Equals A093915 o A000217 = A093916 + A023443. - M. F. Hasler, Apr 04 2009
a(n) = (3+(-1)^n+2*n+(5+(-1)^n)*n^2)/4. a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). G.f.: -x*(x^2+2*x+2)*(x^3-x^2+3*x+1) / ((x-1)^3*(x+1)^3). - Colin Barker, Dec 18 2012

Extensions

Edited and extended beyond a(6) by M. F. Hasler, Apr 04 2009

A093915 Triangle with r-th row containing r consecutive integers that sum to the smallest possible proper multiple of A006003(r).

Original entry on oeis.org

2, 7, 8, 9, 10, 11, 24, 25, 26, 27, 24, 25, 26, 27, 28, 53, 54, 55, 56, 57, 58, 47, 48, 49, 50, 51, 52, 53, 94, 95, 96, 97, 98, 99, 100, 101, 78, 79, 80, 81, 82, 83, 84, 85, 86, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 212, 213, 214, 215, 216, 217
Offset: 1

Views

Author

Amarnath Murthy, Apr 25 2004

Keywords

Comments

The r-th row constructed as explained in the example starts with x=A093916(r), ends with x+r-1=A093918(r), and has its sum A093917(r) equal to the smallest proper multiple of A006003(r). There is a simple formula for A093917(r), which allows us to calculate A093915(n) directly. - M. F. Hasler, Apr 04 2009

Examples

			Given the triangle
1 . . . . with row sum S1 = 1 = A006003(1)
2,3 . . . with row sum S2 = 2+3 = 5 = A006003(2)
4,5,6 . . with row sum S3 = 4+5+6 = 15 = A006003(3), etc.,
the sequence is constructed as follows:
The first row below must be a proper (i.e., > 1) multiple of S1; the smallest possibility is [ 2 ].
The next row below must contain 2 consecutive integers with sum equal to a proper multiple of S2=5. It cannot be 10 (not the sum of 2 consecutive integers), but 15 = 7+8 is a possibility.
The third row [x,x+1,x+2] must sum to a multiple of S3=15, and 2*S3=30 is possible for x=9.
The 4th row [x,x+1,x+2,x+3] must have its sum 4x+6 equal to a multiple of S4=7+8+9+10=34, and x=24 gives the sum 102=3*34, while 2*34=68 can't be achieved for any integer x.
This gives:
2 . . . . . . . with row sum 2 = 2*S1
7,8 . . . . . . with row sum 7+8 = 15 = 3*S2
9,10,11 . . . . with row sum 9+10+11 = 30 = 2*S3
24,25,26,27 . . with row sum 24+25+26+27 = 102 = 3*S4.

Crossrefs

Cf. A093916, A093917, A093918.

Programs

Extensions

Edited and extended (values beyond a(15), example, PARI code) by M. F. Hasler, Apr 04 2009

A093916 a(2*k-1) = (2*k-1)^2 + 2 - k, a(2*k) = 6*k^2 + 2 - k: First column of the triangle A093915.

Original entry on oeis.org

2, 7, 9, 24, 24, 53, 47, 94, 78, 147, 117, 212, 164, 289, 219, 378, 282, 479, 353, 592, 432, 717, 519, 854, 614, 1003, 717, 1164, 828, 1337, 947, 1522, 1074, 1719, 1209, 1928, 1352, 2149, 1503, 2382, 1662, 2627, 1829, 2884, 2004, 3153, 2187, 3434, 2378, 3727, 2577, 4032, 2784, 4349, 2999, 4678, 3222, 5019, 3453, 5372
Offset: 1

Views

Author

Amarnath Murthy, Apr 25 2004

Keywords

Comments

The sequence was initially defined as the first column of the triangle A093915, constructed by trial and error. It is however easy to prove that the sum of the r-th row of A093915, A093917(r), equals twice A006003(r) when r is odd, and three times A006003(r) when r is even. Given the expression of the row sum A093917(r) in terms of the first element a(r), one obtains the explicit formula for a(r). - M. F. Hasler, Apr 04 2009

Links

Crossrefs

Cf. A093915, A093917, A093918.

Programs

  • Magma
    [(n*(5*n-2) + (-1)^n*(n^2+1) + 7)/4: n in [1..70]]; // G. C. Greubel, Dec 30 2021
  • Mathematica
    LinearRecurrence[{0,3,0,-3,0,1},{2,7,9,24,24,53},80] (* Harvey P. Dale, Nov 24 2017 *)
  • PARI
    A093916(n)=((n^2+1)*(3-n%2)-n+1)/2
/* or the "experimental" version, trying out all allowed values */
A093916(n)={ local( s=(n^3+n)/2, d=(n^2-n)/2, k=ceil((2*s-d)/n)); while( (n*k+d)%s, k++ ); k } \\ M. F. Hasler, Apr 04 2009
  • SageMath
    [(5*n^2 -2*n +7 +(-1)^n*(n^2 +1))/4 for n in (1..70)] # G. C. Greubel, Dec 30 2021

Formula

a(n) = ((n^2+1)*b(n) - n + 1)/2 where b(n) = 3 - (n mod 2) = 2 if n odd, = 3 if n even. - M. F. Hasler, Apr 04 2009
From Colin Barker, May 01 2012: (Start)
a(n) = (n*(5*n-2) + (n^2+1)*(-1)^n + 7)/4.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
G.f.: x*(2+7*x+3*x^2+3*x^3+3*x^4+2*x^5)/((1-x)^3*(1+x)^3). (End)
E.g.f.: (1/4)*( (7 +3*x +5*x^2)*exp(x) - 8 + (1 -x +x^2)*exp(-x) ). - G. C. Greubel, Dec 30 2021

Extensions

Edited and extended by M. F. Hasler, Apr 04 2009
Showing 1-3 of 3 results.

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