A048739 -id:A048739 - cp's OEIS frontend

A295380 Number of canonical forms for separation coordinates on hyperspheres S_n, ordered by increasing number of independent continuous parameters.

1, 1, 1, 2, 3, 1, 3, 8, 5, 1, 6, 20, 22, 8, 1, 11, 49, 73, 46, 11, 1, 23, 119, 233, 206, 87, 15, 1, 46, 288, 689, 807, 485, 147, 19, 1, 98, 696, 1988, 2891, 2320, 1021, 236, 24, 1, 207, 1681, 5561, 9737, 9800, 5795, 1960, 356, 29, 1, 451, 4062, 15322, 31350, 38216, 28586, 13088, 3525, 520, 35, 1, 983, 9821, 41558, 97552, 139901, 127465, 74280, 27224, 5989, 730, 41, 1

Offset: 1

Tom Copeland, Nov 21 2017

Table 1 of the Schöbel and Veselov paper with initial 1 added. Reverse of Table 2 of the Devadoss and Read paper.
Apparently A032132 contains the row sums.
From Petros Hadjicostas, Jan 28 2018: (Start)
In this triangle, which is read by rows, for 0 <= k <= n-1 and n>=1, let T(n,k) be the number of inequivalent canonical forms for separation coordinates of the hypersphere S^n with k independent continuous parameters. It is the mirror image of sequence A232206, that is, T(n, k) = A232206(n+1, n-k) for 0 <= k <= n-1 and n>=1. (Triangular array A232206(N, K) is defined for N >= 2 and 1 <= K <= N-1.)
If B(x,y) = Sum_{n,k>=0} T(n,k)*x^n*y^k (with T(0,0) = 1, T(0,k) = 0 for k>=1, and T(n,k) = 0 for 1 <= n <= k), then B(x,y) = 1 + (x/2)*(B(x,y)^2/(1-x*y*B(x,y)) + (1 + x*y*B(x,y))*B(x^2,y^2)/(1-x^2*y^2*B(x^2,y^2))). This can be derived from the bivariate g.f. of A232206. See the comments for that sequence.
Let S(n) := Sum_{k>=0} T(n,k). The g.f. of S(n) is B(x, y=1). If we let y=1 in the above functional equation, we get x*B(x,1) = x + (1/2)*((x*B(x,1))^2/(1-x*B(x,1)) + (1 + x*B(x,1))*x^2*B(x^2,1)/(1-x^2*B(x^2,1))). After some algebra, we get 2*x*B(x,1) = x + (1/2)(x*B(x,1)/(1-x*B(x,1)) + (x*B(x,1) + x^2*B(x^2,1))/(1-x^2*B(x,1))), i.e., 2*x*B(x,1) = x + BIK(x*B(x,1)), where we have the "BIK" (reversible, indistinct, unlabeled) transform of C. G. Bower. This proves that S(n) = A032132(n+1) for n>=0, which is Copeland's claim above.
Note that for the second column we have T(n,k=2) = A048739(n-2) for 2 <= n < = 10, but T(11,2) = 4062 <> 4059 = A048739(9). In any case, they have different g.f.s (see the formula section below).
(End)

			From _Petros Hadjicostas_, Jan 27 2018: (Start)
Triangle T(n,k) begins:
n\k      0     1     2     3     4     5     6    7   8  9
----------------------------------------------------------------
(S^1)    1,
(S^2)    1,    1,
(S^3)    2,    3,    1,
(S^4)    3,    8,    5,    1,
(S^5)    6,   20,   22,    8,    1,
(S^6)   11,   49,   73,   46,   11,    1,
(S^7)   23,  119,  233,  206,   87,   15,    1,
(S^8)   46,  288,  689,  807,  485,  147,   19,   1,
(S^9)   98,  696, 1988, 2891, 2320, 1021,  236,  24,  1,
(S^10) 207, 1681, 5561, 9737, 9800, 5795, 1960, 356, 29, 1,
...
(End)

C. G. Bower, Transforms (2)
S. Devadoss and R. C. Read, Cellular structures determined by polygons and trees, arXiv/0008145 [math.CO], 2000.
S. L. Devadoss and R. C. Read, Cellular structures determined by polygons and trees, Ann. Combin., 5 (2001), 71-98.
K. Schöbel and A. Veselov, Separation coordinates, moduli spaces, and Stasheff polytopes, arXiv:1307.6132 [math.DG], 2014.
K. Schöbel and A. Veselov, Separation coordinates, moduli spaces and Stasheff polytopes, Commun. Math. Phys., 337 (2015), 1255-1274.

Cf. A001190, A024206, A032132, A232206.

From Petros Hadjicostas, Jan 28 2018: (Start)
G.f.: If B(x,y) = Sum_{n,k>=0} T(n,k)*x^n*y^k (with T(0,0) = 1, T(0,k) = 0 for k>=1, and T(n,k) = 0 for 1 <= n <= k), then B(x,y) = 1 + (x/2)*(B(x,y)^2/(1-x*y*B(x,y)) + (1 + x*y*B(x,y))*B(x^2,y^2)/(1-x^2*y^2*B(x^2,y^2))).
If c(N,K) = A232206(N,K) and C(x,y) = Sum_{N,K>=0} c(N,K)*x^N*y^K (with c(1,0) = 1 and c(N,K) = 0 for 0 <= N <= K), then C(x,y) = x*B(x*y, 1/y) and B(x,y) = C(x*y, 1/y)/(x*y).
Setting y=0 in the above functional equation, we get x*B(x,0) = x + (1/2)*((x*B(x,0))^2 + x^2*B(x^2,0)), which is the functional equation for the g.f. of the first column. This proves that T(n,k=0) = A001190(n+1) for n>=0 (assuming T(0,0) = 1).
The g.f. of the second column is B_1(x,0) = Sum_{n>=0} T(n,2)*x^n = lim_{y->0} (B(x,y)-B(x,0))/y, where B(x,0) = 1 + x + x^2 + ... is the g.f. of the first column. We get B_1(x,0) = x*B(x,0)*(B(x,0) - 1)/(1 - x*B(x,0)).
(End)

Typo for T(11,3)=15322 corrected by Petros Hadjicostas, Jan 28 2018

A048773 Partial sums of A048697.

Original entry on oeis.org

1, 11, 32, 84, 209, 511, 1240, 3000, 7249, 17507, 42272, 102060, 246401, 594871, 1436152, 3467184, 8370529, 20208251, 48787040, 117782340, 284351729, 686485807, 1657323352, 4001132520, 9659588401, 23320309331, 56300207072, 135920723484, 328141654049

Offset: 0

Barry E. Williams

Index entries for linear recurrences with constant coefficients, signature (3,-1,-1).

Cf. A001333, A000129, A048696, A048697.

Accumulate[LinearRecurrence[{2,1},{1,10},35]] (* Harvey P. Dale, Jul 26 2011 *)
LinearRecurrence[{3, -1, -1},{1, 11, 32},29] (* Ray Chandler, Aug 03 2015 *)

a(n) = 2*a(n-1)+a(n-2)+9; a(0)=1, a(1)=11.
a(n) = (((10+(11/2)*sqrt(2))*(1+sqrt(2))^n - (10-(11/2)*sqrt(2))*(1-sqrt(2))^n)/ 2*sqrt(2))-9/2.
From R. J. Mathar, Nov 08 2012: (Start)
G.f.: ( 1+8*x ) / ( (x-1)*(x^2+2*x-1) ).
a(n) = A048739(n)+8*A048739(n-1). (End)
a(n) = 3*a(n-1)-a(n-2)-a(n-3). - Wesley Ivan Hurt, May 21 2021

More terms from Harvey P. Dale, Jul 26 2011

A062507 Table by antidiagonals related to partial sums and differences of Pell numbers (A000129).

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 2, 1, 1, 0, 4, 3, 2, 1, 0, 10, 7, 5, 3, 1, 0, 24, 17, 12, 8, 4, 1, 0, 58, 41, 29, 20, 12, 5, 1, 0, 140, 99, 70, 49, 32, 17, 6, 1, 0, 338, 239, 169, 119, 81, 49, 23, 7, 1, 0, 816, 577, 408, 288, 200, 130, 72, 30, 8, 1, 0, 1970, 1393, 985, 696, 488, 330, 202, 102

Offset: 0

Henry Bottomley, Jul 09 2001

			Rows start (0,1,0,2,4,10,...), (0,1,1,3,7,17,...), (0,1,2,5,12,29,...) etc.

Rows are effectively A052542, A001333, A000129, A048739, A048776. Columns are effectively A000004, A000012, A001477, A022856.

T(n, k) =T(n, k-1)+T(n-1, k) =2T(n, k-1)+T(n, k-2)+C(n+k-3, k) for n>2.

A082585 a(1)=1, a(n) = ceiling(r(5)a(n-1)) where r(5) = (1/2)(5 + sqrt(29)) is the positive root of X^2 = 5*X + 1.

Original entry on oeis.org

1, 6, 32, 167, 868, 4508, 23409, 121554, 631180, 3277455, 17018456, 88369736, 458867137, 2382705422, 12372394248, 64244676663, 333595777564, 1732223564484, 8994713599985, 46705791564410, 242523671422036

Offset: 1

Benoit Cloitre, May 07 2003

Index entries for linear recurrences with constant coefficients, signature (6,-4,-1).

Cf. A000071, A048739, A049652, A082574.

a:=n->sum(fibonacci(i,5), i=0..n): seq(a(n), n=1..21); # Zerinvary Lajos, Mar 20 2008

Join[{a=1,b=6},Table[c=5*b+1*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011*)
LinearRecurrence[{6,-4,-1},{1,6,32},30] (* Harvey P. Dale, Jan 06 2012 *)

For n > 3, a(n) = 6*a(n-1) - 4*a(n-2) - a(n-3); a(n) = floor(t(5)*r(5)^n) where t(5) = (1/10)*(1 + 7/sqrt(29)) is the positive root of 145*X^2 = 29*X + 1.
G.f.: x/((x-1)*(x^2+5*x-1)). - Colin Barker, Jan 27 2013
G.f.: 1/(1/Q(0) + 3*x^3 - 3*x) where Q(k) = 1 + k*(2*x+1) + 8*x - 2*x*(k+1)*(k+5)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Mar 15 2013

A128514 Triangle, Pell sequence in every column.

Original entry on oeis.org

1, 2, 1, 5, 2, 1, 12, 5, 2, 1, 29, 12, 5, 2, 1, 70, 29, 12, 5, 2, 1, 169, 70, 29, 12, 5, 2, 1, 408, 169, 70, 29, 12, 5, 2, 1, 985, 408, 169, 70, 29, 12, 5, 2, 1, 2378, 985, 408, 169, 70, 29, 12, 5, 2, 1

Offset: 1

Gary W. Adamson, Mar 05 2007

Row sums = A048739 starting (1, 3, 8, 20, 49, 119, ...).

Riordan array (1/(1-2x-x^2), x). - Philippe Deléham, Apr 23 2009

			First few rows of the triangle:
   1;
   2,  1;
   5,  2,  1;
  12,  5,  2,  1;
  29, 12,  5,  2,  1;
  70, 29, 12,  5,  2,  1;
  ...

Cf. A000129, A048739.

A000129 in every column: (1, 2, 5, 12, 29, ...).

A270359 Positive integer averages of first n Pell numbers; Sum{k=0..n-1} A000129(k) / n where n is in A270342.

Original entry on oeis.org

1, 2, 4, 17, 36, 369, 1820, 20808, 47280, 246561, 6919153, 16008300, 1086517900, 5924129729, 13855173264, 982740019940, 30127233316440, 167427203210673, 5203545562472737, 12300752138736600, 913640750713307860, 162500024938034177361

Offset: 1

Altug Alkan, Mar 15 2016

nonn

			17 is a term because (0 + 1 + 2 + 5 + 12 + 29 + 70) / 7 = 119 / 7 = 17.

Cf. A000129, A048739, A270342.

a048739(n) = local(w=quadgen(8)); -1/2+(3/4+1/2*w)*(1+w)^n+(3/4-1/2*w)*(1-w)^n;
for(n=1, 1e2, if(a048739(n-1) % (n+1) == 0, print1(a048739(n-1) / (n+1) , ", ")));

A270449 Odd integers n such that the sum of the Pell numbers A000129(0) + ... + A000129(n-1) is divisible by n*(n+1)/2.

Original entry on oeis.org

13, 61, 157, 181, 193, 337, 385, 397, 541, 673, 733, 769, 877, 1153, 1201, 1213, 1453, 1873, 1933, 2017, 2029, 2557, 2593, 2797, 3217, 3313, 3517, 4177, 4273, 4561, 4621, 4657, 5101, 5233, 5437, 5581, 5641, 6337, 6637, 6781, 7057, 7213, 7393, 7481, 7537, 7561, 7933, 8221, 8317

Offset: 1

Altug Alkan, Mar 17 2016

nonn

Sequence contains the prime numbers most of the time. Nonprime terms of this sequence are 385, 111361, 111841, 155041, 186961 ...

			13 is a term because (0 + 1 + 2 + 5 + 12 + 29 + 70 + 169 + 408 + 985 + 2378 + 5741 + 13860) / (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13) = 260.

Cf. A000129, A048739, A270342.

a048739(n) = local(w=quadgen(8)); -1/2+(3/4+1/2*w)*(1+w)^n+(3/4-1/2*w)*(1-w)^n;
for(n=1, 1e4, if(a048739(n-1) % ((n+1)*(n+2)/2) == 0 && (n+1) % 2 == 1, print1((n+1), ", ")));

cp's OEIS Frontend

A295380 Number of canonical forms for separation coordinates on hyperspheres S_n, ordered by increasing number of independent continuous parameters.

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A048773 Partial sums of A048697.

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A062507 Table by antidiagonals related to partial sums and differences of Pell numbers (A000129).

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A082585 a(1)=1, a(n) = ceiling(r(5)*a(n-1)) where r(5) = (1/2)*(5 + sqrt(29)) is the positive root of X^2 = 5*X + 1.

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A128514 Triangle, Pell sequence in every column.

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A270359 Positive integer averages of first n Pell numbers; Sum{k=0..n-1} A000129(k) / n where n is in A270342.

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PARI

A270449 Odd integers n such that the sum of the Pell numbers A000129(0) + ... + A000129(n-1) is divisible by n*(n+1)/2.

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PARI

A082585 a(1)=1, a(n) = ceiling(r(5)a(n-1)) where r(5) = (1/2)(5 + sqrt(29)) is the positive root of X^2 = 5*X + 1.