A073762 -id:A073762 - cp's OEIS frontend

A055926 Numbers k such that {largest m such that 1, 2, ..., m divide k} is different from {largest m such that m! divides k}; numbers k which are either odd multiples of 12 or the largest m such that (m-1)! divides k is a composite number > 5.

12, 36, 60, 84, 108, 120, 132, 156, 180, 204, 228, 240, 252, 276, 300, 324, 348, 360, 372, 396, 420, 444, 468, 480, 492, 516, 540, 564, 588, 600, 612, 636, 660, 684, 708, 732, 756, 780, 804, 828, 840, 852, 876, 900, 924, 948, 960, 972, 996, 1020, 1044, 1068

Offset: 1

Leroy Quet, Jul 16 2000

From Antti Karttunen, Nov 20 - Dec 06 2013: (Start)
This sequence has several interpretations:
Numbers k such that A055874(k) differs from A055881(k). [Leroy Quet's original definition of the sequence. Note that A055874(k) >= A055881(k) for all k.]
Numbers k such that {largest m such that m! divides k^2} is different from {largest m such that m! divides k}, i.e., numbers k for which A232098(k) > A055881(k).
Numbers k which are either 12 times an odd number (A073762) or the largest m such that (m-1)! divides k is a composite number > 5 (A232743).
Please see my attached notes for the proof of the equivalence of these interpretations.
Additional implications based on that proof:
A232099 is a subset of this sequence.
A055881(a(n))+1 is always composite. In the range n = 1..17712, only values 4, 6, 8, 9 and 10 occur.
The new definition can be also rephrased by saying that the sequence contains all the positive integers k whose factorial base representation of (A007623(k)) either ends as '...200' (in which case k is an odd multiple of 12, 12 = '200', 36 = '1200', 60 = '2200', ...) or the number of trailing zeros + 2 in that representation is a composite number greater than or equal to 6, e.g. 120 = '10000' (in other words, A055881(k) is one of the terms of A072668 after the initial 3). Together these conditions also imply that all the terms are divisible by 12.
(End)

			12 is included because 3! is the largest factorial to divide 12, but 1, 2, 3 and 4 all divide 12. Equally, 12 is included because it is one of the terms of A073762, or equally, because its factorial base representation ends with digits '...200': A007623(12) = 200.
840 (= 3*5*7*8) is included because the largest factorial which divides 840 is 5! (840 = 7*120), but all positive integers up to 8 divide 840. Equally, 840 is included because it is one of the terms of A232743 as 5+1 = 6 is a composite number larger than 5. Note that A007623(840) = 110000.

Antti Karttunen, Table of n, a(n) for n = 1..17712
Antti Karttunen, A proof for the equivalence of three alternative definitions of A055926
Wikipedia, Wilson's theorem (See the section "Composite modulus")

Union of A073762 and A232743. Equivalently, setwise difference of A232742 and A017593. Subset: A232099.

Cf. A055874, A055881, A072668, A232098, A232100, A232741, A232744, A232745.

More terms from Antti Karttunen, Dec 01 2013

A259748 a(n) = (Sum_{0

Original entry on oeis.org

0, 0, 2, 3, 0, 1, 0, 2, 6, 0, 0, 5, 0, 7, 10, 4, 0, 12, 0, 15, 14, 11, 0, 22, 0, 0, 18, 21, 0, 5, 0, 8, 22, 0, 0, 15, 0, 19, 26, 10, 0, 28, 0, 33, 30, 23, 0, 44, 0, 0, 34, 39, 0, 9, 0, 14, 38, 0, 0, 25, 0, 31, 42, 16, 0, 44, 0, 51, 46, 35, 0, 66, 0, 0, 50

Offset: 1

José María Grau Ribas, Jul 04 2015

{a(n)/n: n=1,2,...} = {0, 1/6, 1/4, 5/12, 1/2, 2/3, 3/4, 11/12}.
From Danny Rorabaugh, Oct 22 2015: (Start)
a(n)/n = 0     iff n mod 24 = 1,2,5,7,10,11,13,17,19,23 (A259749);
a(n)/n = 1/6   iff n mod 24 = 6                         (A259752);
a(n)/n = 1/4   iff n mod 24 = 8,16                      (A259751);
a(n)/n = 5/12  iff n mod 24 = 12                        (A073762);
a(n)/n = 1/2   iff n mod 24 = 14,22                     (A259750);
a(n)/n = 2/3   iff n mod 24 = 3,9,15,18,21              (A259754);
a(n)/n = 3/4   iff n mod 24 = 4,20                      (A259755);
a(n)/n = 11/12 iff n mod 24 = 0                         (A008606).
(End)

Danny Rorabaugh, Table of n, a(n) for n = 1..24000
Danny Rorabaugh, Proof of a(n)/n values for A259748

Cf. A000914,
    A259749 (n such that   a(n)=0),
    A259750 (n such that n/a(n)=2),
    A259751 (n such that n/a(n)=4),
    A259752 (n such that n/a(n)=6),
    A073762 (n such that n/a(n)=12/5),
    A259754 (n such that n/a(n)=3/2),
    A259755 (n such that n/a(n)=4/3),
    A008606 (n such that n/a(n)=12/11).

A[n_]:=Sum[a b,{a,1,n},{b,a+1,n}];Table[Mod[A[n],n],{n,1,122}]

vector(100, n, ((n-1)*n*(n+1)*(3*n+2)/24) % n) \\ Altug Alkan, Oct 22 2015

a(n) = A000914(n) mod n = (1/24)*(-1 + n)*n*(1 + n)*(2 + 3*n) mod n.

a(24k) = 22k; a(24k+1) = 0; a(24k+2) = 0; a(24k+3) = 16k+2; a(24k+4) = 18k+3; a(24k+5) = 0; a(24k+6) = 4k+1, a(24k+7) = 0; a(24k+8) = 6k+2; a(24k+9) = 16k+6; a(24k+10) = 0; a(24k+11) = 0; a(24k+12) = 10k+5; a(24k+13) = 0; a(24k+14) = 12k+7; a(24k+15) = 16k+10; a(24k+16) = 6k+4; a(24k+17) = 0; a(24k+18) = 16k+12; a(24k+19) = 0; a(24k+20) = 18k+15; a(24k+21) = 16k+14; a(24k+22) = 12k+11; a(24k+23) = 0. - Danny Rorabaugh, Oct 22 2015

A259755 Numbers that are congruent to {4, 20} mod 24.

Original entry on oeis.org

4, 20, 28, 44, 52, 68, 76, 92, 100, 116, 124, 140, 148, 164, 172, 188, 196, 212, 220, 236, 244, 260, 268, 284, 292, 308, 316, 332, 340, 356, 364, 380, 388, 404, 412, 428, 436, 452, 460, 476, 484, 500, 508, 524, 532, 548, 556, 572, 580, 596, 604, 620, 628

Offset: 1

José María Grau Ribas, Jul 18 2015

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000914, A007310.

Other sequences of numbers k such that A259748(k)/k equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259754.

[2*(6*n+(-1)^n-3): n in [1..60]]; // Vincenzo Librandi, Aug 27 2015

A[n_] := A[n] = Sum[a b, {a, 1,n}, {b, a + 1, n}]; Select[Range[200], Mod[A[#], #]/# == 3/4 &]
Table[2 (6 n + (-1)^n - 3), {n, 1, 60}] (* Bruno Berselli, Oct 23 2015 *)
LinearRecurrence[{1,1,-1},{4,20,28},60] (* Harvey P. Dale, Jul 19 2016 *)

vector(100, n, 2*(6*n+(-1)^n-3)) \\ Altug Alkan, Oct 23 2015

a(n) = 2*(6*n + (-1)^n - 3).
A259748(a(n))/a(n) = 3/4.
a(n) = 4*A007310(n). - Michel Marcus, Sep 22 2015
G.f.: 4*x*(1 + 4*x + x^2) / ((1 + x)*(1 - x)^2). - Bruno Berselli, Oct 23 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/24. - Amiram Eldar, Dec 31 2021
E.g.f.: 2*(2 + (6*x - 3)*exp(x) + exp(-x)). - David Lovler, Sep 05 2022

Better name from Danny Rorabaugh, Oct 22 2015

A101103 Partial sums of A101104. First differences of A005914.

Original entry on oeis.org

1, 13, 36, 60, 84, 108, 132, 156, 180, 204, 228, 252, 276, 300, 324, 348, 372, 396, 420, 444, 468, 492, 516, 540, 564, 588, 612, 636, 660, 684, 708, 732, 756, 780, 804, 828, 852, 876, 900, 924, 948, 972, 996, 1020, 1044, 1068, 1092, 1116, 1140, 1164, 1188, 1212, 1236, 1260

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

For more information, cross-references etc., see A101104.

For n >= 3, a(n) is equal to the number of functions f:{1,2,3,4}->{1,2,...,n} such that Im(f) contains 3 fixed elements. - Aleksandar M. Janjic and Milan Janjic, Mar 08 2007

G. C. Greubel, Table of n, a(n) for n = 1..5000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq., Vol. 16 (2013) , Article 13.5.7.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A073762.

Concatenation([1,13],List([3..60],n->24*n-36)); # Muniru A Asiru, Dec 02 2018

I:=[36,60]; [1,13] cat [n le 2 select I[n] else 2*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 01 2018

seq(coeff(series(x*(1+x)*(1+10*x+x^2)/(1-x)^2,x,n+1), x, n), n = 1 .. 60); # Muniru A Asiru, Dec 02 2018

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 2, 2}, {k, 0, 34}] OR SeriesAtLevelR = Sum[Eulerian[n, i - 1]*Binomial[n + x - i + r, n + r], {i, 1, n}]; Table[SeriesAtLevelR, {n, 4, 4}, {r, -3, -3}, {x, 3, 35}]
Join[{1, 13},LinearRecurrence[{2, -1},{36, 60},33]] (* Ray Chandler, Sep 23 2015 *)

my(x='x+O('x^60)); Vec(x*(1+x)*(1+10*x+x^2)/(1-x)^2) \\ G. C. Greubel, Dec 01 2018

s=(x*(1+x)*(1+10*x+x^2)/(1-x)^2).series(x, 30); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 01 2018

a(n) = 2*a(n-1) - a(n-2), n > 4.
G.f.: x*(1+x)*(1 + 10*x + x^2)/(1-x)^2.
a(n) = 24*n - 36, n >= 3.
a(n) = Sum_{j=0..n} (-1)^j*binomial(3, j)*(n - j)^4. [Indices shifted, Nov 01 2010]
a(n) = Sum_{i=1..4} A008292(4,i)*binomial(n-i+1,1). [Indices shifted, Nov 01 2010]
Sum_{n>=1} (-1)^(n+1)/a(n) = 157/156 - Pi/48. - Amiram Eldar, Jan 26 2022

Removed redundant information already in A101104. Reduced formulas by expansion of constants - R. J. Mathar, Nov 01 2010

A259749 Numbers that are congruent to {1,2,5,7,10,11,13,17,19,23} mod 24.

Original entry on oeis.org

1, 2, 5, 7, 10, 11, 13, 17, 19, 23, 25, 26, 29, 31, 34, 35, 37, 41, 43, 47, 49, 50, 53, 55, 58, 59, 61, 65, 67, 71, 73, 74, 77, 79, 82, 83, 85, 89, 91, 95, 97, 98, 101, 103, 106, 107, 109, 113, 115, 119, 121, 122, 125, 127, 130, 131, 133, 137, 139, 143, 145

Offset: 1

José María Grau Ribas, Jul 04 2015

Original name: Numbers n such that A259748(n) = 0.

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-2,2,-2,2,-1).

Cf. A000914.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259750, A259751, A259752, A259754, A259755.

A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]  == 0 & ]
Rest@ CoefficientList[Series[x (1 + x^2) (1 + 2 x^2 - x^3 + 2 x^4 - 2 x^5 + 3 x^6 + x^7)/((1 - x)^2*(1 - x + x^2 - x^3 + x^4) (1 + x + x^2 + x^3 + x^4)), {x, 0, 61}], x] (* Michael De Vlieger, Aug 25 2016 *)
Select[Range[150],MemberQ[{1,2,5,7,10,11,13,17,19,23},Mod[#,24]]&] (* or *) LinearRecurrence[{2,-2,2,-2,2,-2,2,-2,2,-1},{1,2,5,7,10,11,13,17,19,23},70] (* Harvey P. Dale, Jan 15 2022 *)

Vec(x*(1+x^2)*(1+2*x^2-x^3+2*x^4-2*x^5+3*x^6+x^7)/((1-x)^2*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)) + O(x^100)) \\ Colin Barker, Aug 25 2016

A259748(a(n)) = Sum_{x*y: x,y in Z/a(n)Z, x<>y} = 0.

G.f.: x*(1+x^2)*(1+2*x^2-x^3+2*x^4-2*x^5+3*x^6+x^7) / ((1-x)^2*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)). - Colin Barker, Aug 25 2016

Better name from Danny Rorabaugh, Oct 22 2015

A259751 Numbers that are congruent to {8, 16} mod 24.

Original entry on oeis.org

8, 16, 32, 40, 56, 64, 80, 88, 104, 112, 128, 136, 152, 160, 176, 184, 200, 208, 224, 232, 248, 256, 272, 280, 296, 304, 320, 328, 344, 352, 368, 376, 392, 400, 416, 424, 440, 448, 464, 472, 488, 496, 512, 520, 536, 544, 560, 568, 584, 592, 608, 616, 632

Offset: 1

José María Grau Ribas, Jul 06 2015

Original name: Numbers n such that n/A259748(n) = 4.

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000914, A001651.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259752, A259754, A259755.

A[n_] := A[n] = Sum[a b, {a, 1,  n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/4 & ]

Vec(8*x*(1+x+x^2)/((1-x)^2*(1+x)) + O(x^100)) \\ Colin Barker, Aug 26 2016

A259748(a(n))/a(n) = 1/4.
a(n) = 8*A001651. - Danny Rorabaugh, Oct 22 2015
From Colin Barker, Aug 26 2016: (Start)
a(n) = 12*n-2*(-1)^n-6.
a(n) = a(n-1)+a(n-2)-a(n-3) for n>3.
G.f.: 8*x*(1+x+x^2) / ((1-x)^2*(1+x)).
(End)
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/72. - Amiram Eldar, Dec 31 2021
E.g.f.: 2*(4 + (6*x - 3)*exp(x) - exp(-x)). - David Lovler, Sep 05 2022

Better name from Danny Rorabaugh, Oct 22 2015

A259752 a(n) = 24*n - 18.

Original entry on oeis.org

6, 30, 54, 78, 102, 126, 150, 174, 198, 222, 246, 270, 294, 318, 342, 366, 390, 414, 438, 462, 486, 510, 534, 558, 582, 606, 630, 654, 678, 702, 726, 750, 774, 798, 822, 846, 870, 894, 918, 942, 966, 990, 1014, 1038, 1062, 1086, 1110, 1134, 1158, 1182, 1206

Offset: 1

José María Grau Ribas, Jul 12 2015

Original name: Numbers n such that n/A259748(n) = 6.

Partial sums give A152746. - Leo Tavares, Jul 29 2023

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000914, A016813, A063128, A063148, A152746.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259751, A259754, A259755.

A[n_] := A[n] = Sum[a b, {a, 1,  n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/6 & ]

A259748(a(n))/a(n) = 1/6.
a(n) = 6*A016813(n-1). - Michel Marcus, Jul 18 2015
G.f.: 6*x*(3*x+1)/(x-1)^2. - Alois P. Heinz, Jul 29 2023
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: 6*(exp(x)*(4*x - 3) + 3).
a(n) = 2*a(n-1) - a(n-2) for n > 2. (End)

Better name from Danny Rorabaugh, Oct 22 2015

A259754 Numbers that are congruent to {3,9,15,18,21} mod 24.

Original entry on oeis.org

3, 9, 15, 18, 21, 27, 33, 39, 42, 45, 51, 57, 63, 66, 69, 75, 81, 87, 90, 93, 99, 105, 111, 114, 117, 123, 129, 135, 138, 141, 147, 153, 159, 162, 165, 171, 177, 183, 186, 189, 195, 201, 207, 210, 213, 219, 225, 231, 234, 237, 243, 249, 255, 258, 261, 267

Offset: 1

José María Grau Ribas, Jul 12 2015

Original name: Numbers n such that n/A259748(n) = 3/2.

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Cf. A000914.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259755.

A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}]; Select[Range[200], Mod[A[#], #]/# == 2/3 &]
Rest@ CoefficientList[Series[3 x (1 + x) (1 + x + x^2 + x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)), {x, 0, 56}], x] (* Michael De Vlieger, Aug 25 2016 *)
LinearRecurrence[{1,0,0,0,1,-1},{3,9,15,18,21,27},60] (* Harvey P. Dale, Aug 30 2016 *)

Vec(3*x*(1+x)*(1+x+x^2+x^4)/((1-x)^2*(1+x+x^2+x^3+x^4)) + O(x^100)) \\ Colin Barker, Aug 25 2016

A259748(a(n))/a(n) = 2/3.
a(n) = 3*A047584(n). - Michel Marcus, Jul 18 2015
From Colin Barker, Aug 25 2016: (Start)
a(n) = a(n-1)+a(n-5)-a(n-6) for n>6.
G.f.: 3*x*(1+x)*(1+x+x^2+x^4) / ((1-x)^2*(1+x+x^2+x^3+x^4)).
(End)

Better name from Danny Rorabaugh, Oct 22 2015

A337923 a(n) is the exponent of the highest power of 2 dividing the n-th Fibonacci number.

Original entry on oeis.org

0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 5, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 6, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 5, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 4, 0, 0, 1

Offset: 1

Amiram Eldar, Jan 29 2021

nonn

			a(1) = 0 since Fibonacci(1) = 1 is odd.
a(6) = 3 since Fibonacci(6) = 8 = 2^3.
a(12) = 4 since Fibonacci(12) = 144 = 2^4 * 3^2.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Tamás Lengyel, The order of the Fibonacci and Lucas numbers, The Fibonacci Quarterly, Vol. 33, No. 3 (1995), pp. 234-239.

Cf. A000045, A007814, A248174.
Cf. A001651, A016945, A017593, A073762, A184985.
Cf. A090740 (sequence without zeros).

a[n_] := IntegerExponent[Fibonacci[n], 2]; Array[a, 100]

def A337923(n): return int(not n%3)+(int(not n%6)<<1) if n%12 else 2+(~n&n-1).bit_length() # Chai Wah Wu, Jul 10 2022

a(n) = A007814(A000045(n)).
The following 4 formulas completely specify the sequence (Lengyel, 1995):
1. a(n) = 0 if n == 1 (mod 3) or n == 2 (mod 3).
2. a(n) = 1 if n == 3 (mod 6).
3. a(n) = 3 if n == 6 (mod 12).
4. a(n) = A007814(n) + 2 if n == 0 (mod 12).
a(A001651(n)) = 0.
a(A016945(n)) = 1.
a(A017593(n)) = 3.
a(A073762(n)) = 4.
The image of this function is A184985, i.e., all the nonnegative integers excluding 2.
Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = 5/6.
a(3*n) = A090740(n), a(3*n+1) = a(3*n+2) = 0. - Joerg Arndt, Mar 01 2023

A256531 First differences of A256530.

Original entry on oeis.org

0, 1, 8, 12, 28, 12, 36, 60, 68, 12, 36, 60, 84, 108, 132, 156, 148, 12, 36, 60, 84, 108, 132, 156, 180, 204, 228, 252, 276, 300, 324, 348, 308, 12, 36, 60, 84, 108, 132, 156, 180, 204, 228, 252, 276, 300, 324, 348, 372, 396, 420, 444, 468, 492, 516, 540, 564, 588, 612, 636, 660, 684, 708, 732, 628, 12, 36, 60, 84, 108

Offset: 0

Omar E. Pol, Apr 21 2015

Number of cells turned ON at n-th stage of cellular automaton of A256530.

Similar to A261695 which shares infinitely many terms.

			With the positive terms written as an irregular triangle in which the row lengths are the terms of A011782 the sequence begins:
1;
8;
12, 28;
12, 36, 60, 68;
12, 36, 60, 84, 108, 132, 156, 148;
12, 36, 60, 84, 108, 132, 156, 180, 204, 228, 252, 276, 300, 324, 348, 308;
...
The terms of the rows that start with 12 are also the initial terms of A073762, except the last term of every row, hence rows converge to A073762.

Paolo Xausa, Table of n, a(n) for n = 0..8191
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Index entries for sequences related to cellular automata

Cf. A011782, A073762, A139251, A147582, A161411, A161415, A256530, A261695.

With[{z=7},Differences[Join[{0,0},Flatten[Array[(2^#-1)^2+12Range[0,2^(#-1)-1]^2&,z]]]]] (* Generates 2^z terms *) (* Paolo Xausa, Nov 15 2023, after Omar E. Pol *)

cp's OEIS Frontend

A055926 Numbers k such that {largest m such that 1, 2, ..., m divide k} is different from {largest m such that m! divides k}; numbers k which are either odd multiples of 12 or the largest m such that (m-1)! divides k is a composite number > 5.

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A259748 a(n) = (Sum_{0

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A259755 Numbers that are congruent to {4, 20} mod 24.

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Magma

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A101103 Partial sums of A101104. First differences of A005914.

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GAP

Magma

Maple

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A259749 Numbers that are congruent to {1,2,5,7,10,11,13,17,19,23} mod 24.

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A259751 Numbers that are congruent to {8, 16} mod 24.

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A259752 a(n) = 24*n - 18.

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