A001045 -id:A001045 - cp's OEIS frontend

A141060 Fourth quadrisection of Jacobsthal numbers A001045: a(n)=16a(n-1)-5.

3, 43, 683, 10923, 174763, 2796203, 44739243, 715827883, 11453246123, 183251937963, 2932031007403, 46912496118443, 750599937895083, 12009599006321323, 192153584101141163, 3074457345618258603, 49191317529892137643

Offset: 0

Paul Curtz, Jul 30 2008

Jacobsthal numbers ending with the decimal digit 3. - Jianing Song, Aug 30 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (17,-16).

The other quadrisections of A001045 are A195156 (first), A139792 (second), and A144864 (third).

[(1/3)*(1+8*16^n): n in [0..25]]; // Vincenzo Librandi, May 25 2011

LinearRecurrence[{17,-16},{3,43},30] (* Harvey P. Dale, Mar 16 2015 *)

a(n)=8*16^n\3+1 \\ Charles R Greathouse IV, May 25 2011

def A141060(n): return ((1<<(n<<2)+3)|1)//3 # Chai Wah Wu, Apr 18 2025

a(n) = A139792(n) + A013776(n).
a(n+1) - a(n) = 10*A013709(n) = 40*A001025(n).
G.f.: (3-8*x)/((1-x)*(1-16*x)). [Colin Barker, Apr 05 2012]
a(0)=3, a(1)=43, a(n)=17*a(n-1)-16*a(n-2). - Harvey P. Dale, Mar 16 2015
From Jianing Song, Aug 30 2022: (Start)
a(n) = A001045(4*n+3).
a(n) = 10*A141032(n) + 3 = 20*A098704(n+1) + 1 = 40*A131865(n-1) + 1 for n >= 1. (End)

A175286 Pisano period of the Jacobsthal sequence A001045 modulo n.

Original entry on oeis.org

1, 1, 6, 2, 4, 6, 6, 2, 18, 4, 10, 6, 12, 6, 12, 2, 8, 18, 18, 4, 6, 10, 22, 6, 20, 12, 54, 6, 28, 12, 10, 2, 30, 8, 12, 18, 36, 18, 12, 4, 20, 6, 14, 10, 36, 22, 46, 6, 42, 20, 24, 12, 52, 54, 20, 6, 18, 28, 58, 12, 60, 10, 18, 2, 12, 30, 66, 8, 66, 12, 70, 18, 18, 36, 60, 18, 30, 12, 78, 4

Offset: 1

R. J. Mathar, Mar 21 2010

nonn

			Reading the sequence 0, 1, 1, 3, 5, 11, 21, ... modulo n=3, we get 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, ... = A088689, which has a period (1, 1, 0, 2, 2, 0) of length a(n=3) = 6.

Eric Weisstein's World of Mathematics, Pisano period.
Wikipedia, Pisano period.

Cf. A001045, A001175, A088689, A175181.

A211893 G.f.: exp( Sum_{n>=1} 3 * Jacobsthal(n)^n * x^n/n ), where Jacobsthal(n) = A001045(n).

Original entry on oeis.org

1, 3, 6, 36, 561, 98211, 43176384, 116622937722, 1022189210900601, 41675008108242048327, 6377839090284322052067558, 4114890941608928235401688095580, 10460015732506081308723488849683574907, 108482611110966450613465001912856742180485969

Offset: 0

Paul D. Hanna, Apr 24 2012

nonn

Given g.f. A(x), note that A(x)^(1/3) is not an integer series.

			G.f.: A(x) = 1 + 3*x + 6*x^2 + 36*x^3 + 561*x^4 + 98211*x^5 + 43176384*x^6 +...
such that
log(A(x))/3 = x + x^2/2 + 3^3*x^3/3 + 5^4*x^4/4 + 11^5*x^5/5 + 21^6*x^6/6 + 43^7*x^7/7 +...+ Jacobsthal(n)^n*x^n/n +...
Jacobsthal numbers begin:
A001045 = [1,1,3,5,11,21,43,85,171,341,683,1365,2731,5461,10923,...].

Cf. A211892, A211894, A211895, A211896, A207971, A207972, A001045.

Cf. A231292 (Jacobsthal(n)^n).

{Jacobsthal(n)=polcoeff(x/(1-x-2*x^2+x*O(x^n)),n)}
{a(n)=polcoeff(exp(sum(k=1, n, 3*Jacobsthal(k)^k*x^k/k)+x*O(x^n)), n)}
for(n=0, 16, print1(a(n), ", "))

G.f.: exp( Sum_{n>=1} (2^n - (-1)^n)^n / 3^(n-1) * x^n/n ).

A211895 G.f.: exp( Sum_{n>=1} 3 * Jacobsthal(n)^3 * x^n/n ), where Jacobsthal(n) = A001045(n).

Original entry on oeis.org

1, 3, 6, 36, 186, 1254, 8208, 57540, 404619, 2913705, 21146694, 155231256, 1147302756, 8538393900, 63879354096, 480212156664, 3624581868297, 27456690186507, 208644709097070, 1589982296208492, 12147079485362406, 93012131704072698, 713676733469348352

Offset: 0

Paul D. Hanna, Apr 25 2012

nonn

Given g.f. A(x), note that A(x)^(1/3) is not an integer series.

			G.f.: A(x) = 1 + 3*x + 6*x^2 + 36*x^3 + 186*x^4 + 1254*x^5 + 8208*x^6 +...
such that
log(A(x))/3 = x + x^2/2 + 3^3*x^3/3 + 5^3*x^4/4 + 11^3*x^5/5 + 21^3*x^6/6 + 43^3*x^7/7 +...+ Jacobsthal(n)^3*x^n/n +...
Jacobsthal numbers begin:
A001045 = [1,1,3,5,11,21,43,85,171,341,683,1365,2731,5461,10923,...].

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

Cf. A211893, A211894, A211896, A207970, A001045 (Jacobsthal).

CoefficientList[Series[((1+x)*(1+4*x)^3/((1-2*x)^3*(1-8*x)))^(1/9), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 24 2012 *)

{Jacobsthal(n)=polcoeff(x/(1-x-2*x^2+x*O(x^n)),n)}
{a(n)=polcoeff(exp(sum(k=1, n, 3*Jacobsthal(k)^3*x^k/k)+x*O(x^n)), n)}
for(n=0, 30, print1(a(n), ", "))

{a(n)=polcoeff(( (1+x)*(1+4*x)^3 / ((1-2*x)^3*(1-8*x)+x*O(x^n)) )^(1/9),n)}

G.f.: ( (1+x)*(1+4*x)^3 / ((1-2*x)^3*(1-8*x)) )^(1/9).
G.f.: exp( Sum_{n>=1} (2^n - (-1)^n)^3 / 9 * x^n/n ).
Recurrence: n*a(n) = (5*n-2)*a(n-1) + 6*(5*n-12)*a(n-2) - 8*(5*n-12)*a(n-3) - 64*(n-4)*a(n-4). - Vaclav Kotesovec, Oct 24 2012
a(n) ~ 3^(2/9)*8^n/(Gamma(1/9)*n^(8/9)). - Vaclav Kotesovec, Oct 24 2012

A293434 a(n) is the sum of the proper divisors of n that are Jacobsthal numbers (A001045).

Original entry on oeis.org

0, 1, 1, 1, 1, 4, 1, 1, 4, 6, 1, 4, 1, 1, 9, 1, 1, 4, 1, 6, 4, 12, 1, 4, 6, 1, 4, 1, 1, 9, 1, 1, 15, 1, 6, 4, 1, 1, 4, 6, 1, 25, 1, 12, 9, 1, 1, 4, 1, 6, 4, 1, 1, 4, 17, 1, 4, 1, 1, 9, 1, 1, 25, 1, 6, 15, 1, 1, 4, 6, 1, 4, 1, 1, 9, 1, 12, 4, 1, 6, 4, 1, 1, 25, 6, 44, 4, 12, 1, 9, 1, 1, 4, 1, 6, 4, 1, 1, 15, 6, 1, 4, 1, 1, 30

Offset: 1

Antti Karttunen, Oct 09 2017

nonn

			For n = 15, whose proper divisors are [1, 3, 5], all of them are in A001045, thus a(15) = 1 + 3 + 5 = 9.
For n = 21, whose proper divisors are [1, 3, 7], both 1 and 3 are in A001045, thus a(21) = 1 + 3 = 4.
For n = 21845, whose proper divisors are [1, 5, 17, 85, 257, 1285, 4369], only 1, 5, 85 are in A001045, thus a(21845) = 1 + 5 + 85 = 91.

Antti Karttunen, Table of n, a(n) for n = 1..21845
Index entries for sequences related to sums of divisors

Cf. A001045, A147612, A293432, A293433.

Cf. also A293436.

With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Table[DivisorSum[n, # &, And[MemberQ[s, #], # != n] &], {n, 105}]] (* Michael De Vlieger, Oct 09 2017 *)

A147612aux(n,i) = if(!(n%2),n,A147612aux((n+i)/2,-i));
A147612(n) = 0^(A147612aux(n,1)*A147612aux(n,-1));
A293434(n) = sumdiv(n,d,(dA147612(d)*d);

from sympy import divisors
def A293434(n): return sum(d for d in divisors(n,generator=True) if d(m-3).bit_length()) # Chai Wah Wu, Apr 18 2025

a(n) = Sum_{d|n, dA147612(d)*d.

a(n) = A293432(n) - (A147612(n)*n).

A066488 Composite numbers k which divide A001045(k-1).

Original entry on oeis.org

341, 1105, 1387, 1729, 2047, 2465, 2701, 2821, 3277, 4033, 4369, 4681, 5461, 6601, 7957, 8321, 8911, 10261, 10585, 11305, 13741, 13747, 13981, 14491, 15709, 15841, 16705, 18721, 19951, 23377, 29341, 30121, 30889, 31417, 31609, 31621, 34945

Offset: 1

Robert G. Wilson v, Jan 03 2002

Also composite numbers k such that (2^k - 2)/3 + 1 == 2^k - 1 == 1 (mod k).
An equivalent definition of this sequence: pseudoprimes to base 2 that are not divisible by 3. - Arkadiusz Wesolowski, Nov 15 2011
Conjecture: these are composites k such that 2^M(k-1) == -1 (mod M(k)^2 + M(k) + 1), where M(k) = 2^k - 1. - Amiram Eldar and Thomas Ordowski, Dec 19 2019
These are composites k such that 2^(m-1) == 1 (mod (m+1)^6 - 1), where m = 2^k - 1. - Thomas Ordowski, Sep 17 2023

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A001567, A002808, A001045, A066047.

[k:k in [4..40000]|not IsPrime(k) and ((2^(k-1) + (-1)^k) div 3) mod k eq 0]; // Marius A. Burtea, Dec 20 2019

a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n - 1] + 2a[n - 2]; Select[ Range[50000], IntegerQ[a[ # - 1]/ # ] && !PrimeQ[ # ] && # != 1 & ]
fQ[n_] := ! PrimeQ@ n && Mod[((2^n - 2)/3 + 1), n] == Mod[2^n - 1, n] == 1; Select[ Range@ 35000, fQ]

is(n)=n%3 && Mod(2,n)^(n-1)==1 && !isprime(n) && n>1 \\ Charles R Greathouse IV, Sep 18 2013

A123641 Triangular array related to sequence A123640 with row sum A001045.

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 0, 1, 0, 4, 1, 0, 2, 0, 8, 0, 1, 0, 4, 0, 16, 1, 0, 2, 0, 8, 0, 32, 0, 1, 0, 4, 0, 16, 0, 64, 1, 0, 2, 0, 8, 0, 32, 0, 128, 0, 1, 0, 4, 0, 16, 0, 64, 0, 256, 1, 0, 2, 0, 8, 0, 32, 0, 128, 0, 512, 0, 1, 0, 4, 0, 16, 0, 64, 0, 256, 0, 1024, 1

Offset: 1

Alford Arnold, Oct 04 2006

			For n>0 A123640 begins
1
0 1
1 0 1 1
0 1 0 0 1 1 1 1
suppressing multiple zeros and counting contiguous ones yields
1
0 1
1 0 2
0 1 0 4
1 0 2 0 8
0 1 0 4 0 16
...

Cf. A000079, A001045, A059841, A123640.

a001045(n)=(2^n-(-1)^n)/3;
lista(nn) = {last = v[2]; nb = 1; s = 0; j = 1; for (i=3, nn, if ((v[i] == last) && (s+v[i] < a001045(j)), nb++; s += v[i];, if (last, w = nb, w = 0); print1(w, ", "); nb = 1; last = v[i]; if (s + v[i] >= a001045(j), s = 0; j++;, s += v[i];);););} \\ Michel Marcus, Feb 09 2014

More terms from Michel Marcus, Feb 09 2014

A129738 List of primitive prime divisors of the Jacobsthal numbers A001045 in their order of occurrence.

Original entry on oeis.org

3, 5, 11, 7, 43, 17, 19, 31, 683, 13, 2731, 127, 331, 257, 43691, 73, 174763, 41, 5419, 23, 89, 2796203, 241, 251, 4051, 8191, 87211, 29, 113, 59, 3033169, 151, 715827883, 65537, 67, 20857, 131071, 281, 86171, 37, 109, 1777, 25781083, 524287, 22366891, 61681, 83

Offset: 1

N. J. A. Sloane, May 13 2007

nonn

Read A001045 term-by-term, factorize each term, write down any primes not seen before.

Amiram Eldar, Table of n, a(n) for n = 1..3915
Graham Everest, Shaun Stevens, Duncan Tamsett and Tom Ward, Primes generated by recurrence sequences, Amer. Math. Monthly, 114 (No. 5, 2007), 417-431.
K. Zsigmondy, Zur Theorie der Potenzreste, Monatsh. Math., 3 (1892), 265-284.

Cf. A001045, A049883, A107036, A129733.

concat := (a,h)->[op(a),op(sort(convert(h,list)))]:
PPDinOrder := proc(S) local A,H,T,s;
T := {0,1}; A := [];
for s in S do
  H := numtheory[factorset](s) minus T:
  if H <> {} then
    A := concat(A,H);
    T := T union H
  fi
od;
A end:
A129738 := PPDinOrder(A001045);
# Peter Luschny, Jan 04 2011

DeleteDuplicates[Flatten[FactorInteger[#][[All,1]]&/@LinearRecurrence[ {1,2},{3,5},50]]](* Harvey P. Dale, Apr 14 2020 *)

A140944 Triangle T(n,k) read by rows, the k-th term of the n-th differences of the Jacobsthal sequence A001045.

Original entry on oeis.org

0, 1, 0, -1, 2, 0, 3, -2, 4, 0, -5, 6, -4, 8, 0, 11, -10, 12, -8, 16, 0, -21, 22, -20, 24, -16, 32, 0, 43, -42, 44, -40, 48, -32, 64, 0, -85, 86, -84, 88, -80, 96, -64, 128, 0, 171, -170, 172, -168, 176, -160, 192, -128, 256, 0, -341, 342, -340, 344, -336, 352, -320, 384, -256, 512, 0

Offset: 0

Paul Curtz, Jul 24 2008

A variant of the triangle A140503, now including the diagonal.

Since the diagonal contains zeros, rows sums are those of A140503.

			Triangle begins as:
    0;
    1,   0;
   -1,   2,   0;
    3,  -2,   4,  0;
   -5,   6,  -4,  8,   0;
   11, -10,  12, -8,  16,  0;
  -21,  22, -20, 24, -16, 32,  0;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000079, A001045, A016131, A045883, A052992, A083086, A084175.

Cf. A091056, A097038, A109765, A115489, A140503, A192382, A246036.

[2^k*(1-(-2)^(n-k))/3: k in [0..n], n in [0..15]]; // G. C. Greubel, Feb 18 2023

A001045:= n -> (2^n-(-1)^n)/3;
A140944:= proc(n,k) if n = 0 then A001045(k); else procname(n-1,k+1)-procname(n-1,k) ; fi; end:
seq(seq(A140944(n,k),k=0..n),n=0..10); # R. J. Mathar, Sep 07 2009

T[0, 0]=0; T[1, 0]= T[0, 1]= 1; T[0, k_]:= T[0, k]= T[0, k-1] + 2*T[0, k-2]; T[n_, n_]=0; T[n_, k_]:= T[n, k] = T[n-1, k+1] - T[n-1, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Jean-François Alcover, Dec 17 2014 *)
Table[2^k*(1-(-2)^(n-k))/3, {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2023 *)

T(n, k) = (2^k - 2^n*(-1)^(n+k))/3 \\ Jianing Song, Aug 11 2022

def A140944(n,k): return 2^k*(1 - (-2)^(n-k))/3
flatten([[A140944(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Feb 18 2023

T(n, k) = T(n-1, k+1) - T(n-1, k). T(0, k) = A001045(k).
T(n, k) = (2^k - 2^n*(-1)^(n+k))/3, for n >= k >= 0. - Jianing Song, Aug 11 2022
From G. C. Greubel, Feb 18 2023: (Start)
T(n, n-1) = A000079(n).
T(2*n, n) = (-1)^(n+1)*A192382(n+1).
T(2*n, n-1) = (-1)^n*A246036(n-1).
T(2*n, n+1) = A083086(n).
T(3*n, n) = -A115489(n).
Sum_{k=0..n} T(n, k) = A052992(n)*[n>0] + 0*[n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = A045883(n).
Sum_{k=0..n} 2^k*T(n, k) = A084175(n).
Sum_{k=0..n} (-2)^k*T(n, k) = (-1)^(n+1)*A109765(n).
Sum_{k=0..n} 3^k*T(n, k) = A091056(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = (-1)^(n+1)*A097038(n).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = (-1)^(n+1)*A138495(n). (End)

Edited and extended by R. J. Mathar, Sep 07 2009

A200375 Product of Catalan and Jacobsthal numbers: a(n) = A000108(n)*A001045(n+1).

Original entry on oeis.org

1, 1, 6, 25, 154, 882, 5676, 36465, 244530, 1657942, 11471668, 80242890, 568080772, 4056976900, 29212908120, 211783889025, 1544811959970, 11328491394990, 83473572128100, 617702666484750, 4588654943721420, 34206312386929020, 255803818897858920, 1918528298674328250, 14427334095935095764

Offset: 0

Paul D. Hanna, Nov 16 2011

nonn

More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.

			G.f.: A(x) = 1 + x + 2*3*x^2 + 5*5*x^3 + 14*11*x^4 + 42*21*x^5 + 132*43*x^6 + 429*85*x^7 + 1430*171*x^8 +...+ A000108(n)*A001045(n)*x^n +...
The g.f. of the Jacobsthal sequence A001045, F(x) = 1/(1-x-2*x^2), begins:
F(x) = 1 + x + 3*x^2 + 5*x^3 + 11*x^4 + 21*x^5 + 43*x^6 + 85*x^7 + 171*x^8 +...
The g.f. of A200376, where G(x) =  A(x/G(x)), begins:
G(x) = 1 + x + 5*x^2 + 9*x^3 + 37*x^4 + 81*x^5 + 301*x^6 + 729*x^7 +...
in which the odd-indexed coefficients are powers of 9.

Michael De Vlieger, Table of n, a(n) for n = 0..1112
Paul Barry, On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths, arXiv:2101.10218 [math.CO], 2021.
Paul Barry and Arnauld Mesinga Mwafise, Classical and Semi-Classical Orthogonal Polynomials Defined by Riordan Arrays, and Their Moment Sequences, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.5.
S. B. Ekhad and M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017).

Cf. A200376, A098614, A098616, A200312.

Array[CatalanNumber[# - 1] (2^# - (-1)^#)/3 &, 25] (* Michael De Vlieger, Apr 24 2018 *)

{a(n) = binomial(2*n, n)/(n+1) * (2^(n+1) + (-1)^n)/3}

{a(n) = polcoef(sqrt((1-2*x - sqrt(1-4*x-32*x^2 +O(x^(n+3))))/2)/(3*x), n)}

{a(n) = polcoef((1/x)*serreverse(x-x^2 - 4*x^3*sum(m=0,n\2,binomial(2*m,m)/(m+1)*3^m*x^(2*m)) +x^3*O(x^n)), n)}

G.f.: sqrt( (1-2*x - sqrt(1-4*x-32*x^2))/2 )/(3*x).
G.f.: (1/x)*Series_Reversion(x-x^2 - 4*x^3*Sum_{n>=0} A000108(n)*3^n*x^(2*n) ).
G.f. satisfies: A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where G(x) is the g.f. of A200376: G(x) = 1/sqrt(1-10*x^2 + x^4/(1-8*x^2)) + x/(1-9*x^2).
n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -8*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 17 2011
a(n) = binomial(2*n,n)/(n+1) * (2^(n+1) + (-1)^n)/3.
From Peter Bala, Aug 17 2021: (Start)
G.f.: A(x) = (sqrt(1 + 4*x) - sqrt(1 - 8*x))/(6*x).
A(x) = 1/sqrt(1 + 4*x)*c( 3*x/(1 + 4*x) ), where c(x) = (1 - sqrt(1- 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. Cf. A151374.
In general, [x^n] ( 1/sqrt(1 + 4*x)*c( k*x/(1 + 4*x) ) ) = Catalan(n)*((k-1)^(n+1) + (-1)^(n+1))/k.
A(x) = 1/sqrt(1 - 8*x)*c( -3*x/(1 - 8*x) ). (End)
G.f. A(x) satisfies A(x) = sqrt( 1 + 2*x*A(x)^2 + 9*x^2*A(x)^4 ). - Paul D. Hanna, Dec 14 2024

Typo in Name corrected by Peter Bala, Aug 17 2021

cp's OEIS Frontend

A141060 Fourth quadrisection of Jacobsthal numbers A001045: a(n)=16a(n-1)-5.

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A175286 Pisano period of the Jacobsthal sequence A001045 modulo n.

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A211893 G.f.: exp( Sum_{n>=1} 3 * Jacobsthal(n)^n * x^n/n ), where Jacobsthal(n) = A001045(n).

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A211895 G.f.: exp( Sum_{n>=1} 3 * Jacobsthal(n)^3 * x^n/n ), where Jacobsthal(n) = A001045(n).

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A293434 a(n) is the sum of the proper divisors of n that are Jacobsthal numbers (A001045).

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A066488 Composite numbers k which divide A001045(k-1).

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A123641 Triangular array related to sequence A123640 with row sum A001045.

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A129738 List of primitive prime divisors of the Jacobsthal numbers A001045 in their order of occurrence.

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