A033889 -id:A033889 - cp's OEIS frontend

A154623 Sequence with g.f. 1+(x/(1-5x))c(x/(1-5*x)), c(x) the g.f. of A000108.

1, 1, 6, 37, 235, 1539, 10392, 72267, 516474, 3783115, 28317562, 215969271, 1673702191, 13148444197, 104494340880, 838670818365, 6788255966595, 55346471893395, 454123503938490, 3746885525588175, 31066887028255065

Offset: 0

Paul Barry, Jan 13 2009

Hankel transform is F(4n+1) (A033889).

a(n+1) is the 5th binomial transform of A000108.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

CoefficientList[Series[1/2*(3-Sqrt[(1-9*x)/(1-5*x)]), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

G.f.: (1/2)*(3-sqrt((1-9*x)/(1-5*x))).
a(n)=(4/5)*0^n+sum{k=0..n-1, C(n-1,k)*A000108(k)*5^(n-k-1)}.
Conjecture: n*a(n) +2*(8-7n)*a(n-1) +45*(n-2)*a(n-2) = 0. - R. J. Mathar, Dec 14 2011
a(n) ~ 3^(2*n+1)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
a(n) = (-1)^(n-1)*(GegenbauerC(n-1,-n+1,7/2) + GegenbauerC(n-2,-n+1,7/2)) for n>=1. - Peter Luschny, May 13 2016

A374602 Array of successive integer solutions to sqrt((d-c)b^2 + c(b+1)^2) for nonsquare integers d >= 2 (d=A000037(n) for n >= 1), where b and c are positive integers and c < d, read by antidiagonals.

Original entry on oeis.org

5, 29, 3, 169, 11, 5, 985, 41, 13, 3, 5741, 153, 34, 7, 4, 33461, 571, 89, 18, 5, 10, 195025, 2131, 233, 29, 11, 11, 4, 1136689, 7953, 610, 69, 28, 23, 5, 7, 6625109, 29681, 1597, 178, 62, 58, 13, 8, 6, 38613965, 110771, 4181, 287, 79, 338, 14, 13, 22, 4

Offset: 1

Charles L. Hohn, Jul 13 2024

T(n,k) is the diagonal lengths of increasingly nearly regular d-dimensional Pythagorean hyperrectangles.
Each row n divides into equal length, geometrically periodic subsequences, each with its own subsequence period length (A377290) and geometric growth factor (A377291); it is conjectured that this is the case for all n, and that all solutions conform as such and that there are no solutions that do not, but these are not proven.
It is also not known if there is an algorithm for generating values for all rows other than testing all possible values for a row until a subsequence pattern emerges.
Square d produce solutions following a different pattern, shown as A375336.

			n=row index; d=nonsquare integer of index n (A000037(n)):
 n    d   T(n,k)
---+----+-------------------------------------------------------------
 1 |  2 |  5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, ...
 2 |  3 |  3, 11,  41, 153,  571,  2131,   7953,   29681,  110771, ...
 3 |  5 |  5, 13,  34,  89,  233,   610,   1597,    4181,   10946, ...
 4 |  6 |  3,  7,  18,  29,   69,   178,    287,     683,    1762, ...
 5 |  7 |  4,  5,  11,  28,   62,    79,    175,     446,     988, ...
 6 |  8 | 10, 11,  23,  58,  338,   373,    781,    1970,   11482, ...
 7 | 10 |  4,  5,  13,  14,   25,    62,    111,     148,     185, ...
 8 | 11 |  7,  8,  13,  32,   57,   139,    158,     259,     638, ...
 9 | 12 |  6, 22,  39,  69,   82,   125,    306,     543,    1142, ...
10 | 13 |  4,  5,   7,  17,   30,    43,     53,      76,     185, ...
11 | 14 |  9, 11,  14,  19,   46,    81,    267,     329,     418, ...
12 | 15 |  6, 10,  21,  23,   30,    39,     94,     165,     362, ...
13 | 17 | 25, 27,  34,  41,   98,   171,    260,    1649,    1779, ...
14 | 18 |  6, 13,  15,  18,   21,    50,     87,     132,     198, ...
15 | 19 |  5,  7,   8,   9,   11,    31,     34,      37,      56, ...
16 | 20 | 10, 26,  68, 125,  159,   178,    197,     466,     807, ...
17 | 21 |  6,  9,  12,  13,   14,    33,     57,      86,     134, ...
18 | 22 |  5,  7,   8,  17,   18,    19,     31,      64,      77, ...
19 | 23 | 16, 19,  27,  28,   29,    68,    117,     176,     764, ...
20 | 24 |  6,  9,  11,  14,   36,    39,     57,      58,      59, ...
...
sqrt((2-1)*1^2 + 1*(1+1)^2) = sqrt(5) -> not an integer so not included.
sqrt((2-1)*3^2 + 1*(3+1)^2) = 5 -> T(1,1).
sqrt((2-1)*20^2 + 1*(20+1)^2) = 29 -> T(1,2).
sqrt((3-2)*1^2 + 2*(1+1)^2) = 3 -> T(2,1).
sqrt((6-2)*7^2 + 2*(7+1)^2) = 18 -> T(4,3).

Row 1 is A001653 starting at n=2.
Row 2 is A079935 starting at n=2.
Bisection of row 2 starting with the first term is A189356 starting at n=1.
Bisection of row 2 starting with the second term is A122769 starting at n=2.
Row 3 is A001519 starting at n=3.
Bisection of row 3 starting with the first term is A033889 starting at n=1.
Bisection of row 3 starting with the second term is A033891 starting at n=1.
Row 4 is A131093 starting at n=3.
Cf. A000037, A373666.
Cf. A377290, A377291, A375336.

row(n, c)=my(v=List(), d=n+floor(sqrt(n)+1/2) /* d=A000037(n) */, t=ceil(sqrt(d))); while(#v

T(n, 1) = A373666(A000037(n)).

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

A276529 a(n) = (a(n-1) * a(n-5) + 1) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 13, 20, 27, 34, 41, 89, 137, 185, 233, 281, 610, 939, 1268, 1597, 1926, 4181, 6436, 8691, 10946, 13201, 28657, 44113, 59569, 75025, 90481, 196418, 302355, 408292, 514229, 620166, 1346269, 2072372, 2798475, 3524578, 4250681, 9227465

Offset: 0

Seiichi Manyama, Nov 16 2016

Thanks to the linear recurrence signature, we see that this is actually five separate linear recurrence sequences, each with signature (7,-1), interwoven together. - Greg Dresden, Oct 16 2021

Seiichi Manyama, Table of n, a(n) for n = 0..5985
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,7,0,0,0,0,-1).

Cf. A275173, A102276, A276530.

5th-sections: A049685, A033891, A033889.

LinearRecurrence[{0,0,0,0,7,0,0,0,0,-1}, {1,1,1,1,1,1,2,3,4,5}, 50] (* G. C. Greubel, Nov 18 2016 *)
RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==a[4]==a[5]==1,a[n]==(a[n-1]a[n-5]+ 1)/a[n-6]},a,{n,50}] (* Harvey P. Dale, Oct 08 2020 *)
Flatten[Table[{LucasL[4 n - 2]/3, Fibonacci[4 n - 1], LucasL[4 n + 2]/3 - Fibonacci[4 n], LucasL[4 n - 2]/3 + Fibonacci[4 n], Fibonacci[4 n + 1]}, {n, 0, 10}]] (* Greg Dresden, Oct 16 2021 *)

Vec((1 +x +x^2 +x^3 +x^4 -6*x^5 -5*x^6 -4*x^7 -3*x^8 -2*x^9)/(1 -7*x^5 +x^10) + O(x^50)) \\ Colin Barker, Nov 16 2016

def A(k, m, n)
  a = Array.new(2 * k, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[-1] * a[1] + a[k] ** m
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276529(n)
  A(3, 0, n)
end

a(n) + a(n+10) = 7*a(n+5).
a(5-n) = a(n).
G.f.: (1 +x +x^2 +x^3 +x^4 -6*x^5 -5*x^6 -4*x^7 -3*x^8 -2*x^9) / (1 -7*x^5 +x^10). - Colin Barker, Nov 16 2016
From Greg Dresden, Oct 16 2021: (Start)
a(5*n) = L(4*n-2)/3 = A049685(n-1),
a(5*n+1) = F(4*n-1) = A033891(n-1),
a(5*n+2) = L(4*n+2)/3 - F(4*n),
a(5*n+3) = L(4*n-2)/3 + F(4*n),
a(5*n+4) = F(4*n+1) = A033889(n). (End)

A164698 Semiprimes pq such that pq - 1 divides p^2 + q^2 + 2.

Original entry on oeis.org

6, 21, 26, 51, 1157, 372101, 1288005205276048901

Offset: 1

Mohamed Bouhamida, Aug 22 2009

Semiprimes pq such that pq-1 divides (p+q)^2.
The third to fifth terms are Fib(3)*Fib(7), Fib(7)*Fib(11) and Fib(13)*Fib(17).
Products of two prime Fibonacci numbers F(k) and F(k+4) (see A001605 and A005478) are in the sequence.
6 and 26 are the only even terms. Odd terms contain products of pairs of consecutive terms from the following sequences: A005248, A001541, A033889, A033891. - Max Alekseyev, Aug 27 2009

			The semiprime 6 = 2*3 is in the sequence because 2*3 - 1 = 5 divides 2^2 + 3^2 + 2 = 15.

Cf. A001358, A000045, A140362, A164643.

isA001358 := proc(n) RETURN ( numtheory[bigomega](n) =2 ) ; end:
isA164698 := proc(n) if isA001358(n) then p := op(1,op(1,ifactors(n)[2]) ) ; q := n/p ; if (p^2+q^2+2) mod (p*q-1) = 0 then true; else false; fi; else false; fi; end:
for n from 4 to 3000000 do if isA164698(n) then print(n, ifactors(n)) ; fi; od: # R. J. Mathar, Aug 24 2009

Missing values added by R. J. Mathar, Aug 24 2009

a(7) from Max Alekseyev, Aug 27 2009

cp's OEIS Frontend

A154623 Sequence with g.f. 1+(x/(1-5*x))*c(x/(1-5*x)), c(x) the g.f. of A000108.

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A374602 Array of successive integer solutions to sqrt((d-c)*b^2 + c*(b+1)^2) for nonsquare integers d >= 2 (d=A000037(n) for n >= 1), where b and c are positive integers and c < d, read by antidiagonals.

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A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

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A276529 a(n) = (a(n-1) * a(n-5) + 1) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

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A164698 Semiprimes pq such that pq - 1 divides p^2 + q^2 + 2.

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A154623 Sequence with g.f. 1+(x/(1-5x))c(x/(1-5*x)), c(x) the g.f. of A000108.

A374602 Array of successive integer solutions to sqrt((d-c)b^2 + c(b+1)^2) for nonsquare integers d >= 2 (d=A000037(n) for n >= 1), where b and c are positive integers and c < d, read by antidiagonals.