A017533 -id:A017533 - cp's OEIS frontend

A017535 a(n) = (12*n+1)^3.

1, 2197, 15625, 50653, 117649, 226981, 389017, 614125, 912673, 1295029, 1771561, 2352637, 3048625, 3869893, 4826809, 5929741, 7189057, 8615125, 10218313, 12008989, 13997521, 16194277, 18609625, 21253933, 24137569, 27270901, 30664297, 34328125, 38272753

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000578, A017533.

[(12*n + 1)^3: n in [0..40]]; // Vincenzo Librandi, May 26 2016

Table[(12 n + 1)^3, {n, 0, 50}] (* Vincenzo Librandi, May 26 2016 *)

makelist( (12*n+1)^3  ,n,0,30); /* Martin Ettl, Oct 21 2012 */

G.f.: (1 + 2193*x + 6843*x^2 + 1331*x^3)/(1-x)^4. - Vincenzo Librandi, May 26 2016
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), for n>3. - Vincenzo Librandi, May 26 2016
a(n) = A000578(A017533(n)). - Michel Marcus, May 26 2016

A017540 a(n) = (12*n + 1)^8.

Original entry on oeis.org

1, 815730721, 152587890625, 3512479453921, 33232930569601, 191707312997281, 806460091894081, 2724905250390625, 7837433594376961, 19925626416901921, 45949729863572161, 97906861202319841

Offset: 0

N. J. A. Sloane

Stefano Spezia, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A001016, A017533.

(12*Range[0,20]+1)^8 (* Harvey P. Dale, Jul 23 2013 *)

a(n) = A001016(A017533(n)). - Michel Marcus, Jul 29 2022

A017543 a(n) = (12*n + 1)^11.

Original entry on oeis.org

1, 1792160394037, 2384185791015625, 177917621779460413, 3909821048582988049, 43513917611435838661, 313726685568359708377, 1673432436896142578125, 7153014030880804126753, 25804264053054077850709, 81402749386839761113321, 230339304218442143770717

Offset: 0

N. J. A. Sloane

Matthew House, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (12, -66, 220, -495, 792, -924, 792, -495, 220, -66, 12, -1).

Cf. A008455 (n^11), A017533 (12n+1).

[(12*n+1)^11: n in [0..12]]; // Vincenzo Librandi, Jul 29 2015

seq((12*n+1)^11, n=0..100); # Robert Israel, Jul 28 2015

vector(20, n, n--; (12*n+1)^11) \\ Michel Marcus, Jul 28 2015

a(n) = A008455(A017533(n)). - Michel Marcus, Jul 28 2015

a(n) = Sum_{k=0..11} binomial(11,k)*(12*n)^k. - Robert Israel, Jul 28 2015

A033576 a(n) = (2n+1)(12*n+1).

Original entry on oeis.org

1, 39, 125, 259, 441, 671, 949, 1275, 1649, 2071, 2541, 3059, 3625, 4239, 4901, 5611, 6369, 7175, 8029, 8931, 9881, 10879, 11925, 13019, 14161, 15351, 16589, 17875, 19209, 20591, 22021, 23499, 25025

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005408, A017533.

List([0..50], n-> (2*n+1)*(12*n+1)); # G. C. Greubel, Oct 12 2019

[(2*n+1)*(12*n+1): n in [0..50]]; // Vincenzo Librandi, Jul 07 2012

A033576:=n->(2*n+1)*(12*n+1); seq(A033576(n), n=0..50); # Wesley Ivan Hurt, Feb 02 2014

Table[(2n+1)(12n+1),{n,0,50}]  (* Harvey P. Dale, Mar 30 2011 *)
CoefficientList[Series[(1+36*x+11*x^2)/(1-x)^3,{x,0,50}],x] (* Vincenzo Librandi, Jul 07 2012 *)

a(n)=(2*n+1)*(12*n+1) \\ Charles R Greathouse IV, Jun 17 2017

[(2*n+1)*(12*n+1) for n in range(50)] # G. C. Greubel, Oct 12 2019

From Colin Barker, Jun 10 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (1 + 36*x + 11*x^2)/(1-x)^3. (End)
a(n) = A005408(n) * A017533(n). - Wesley Ivan Hurt, Feb 02 2014
E.g.f.: (1 + 38*x + 24*x^2)*exp(x). - G. C. Greubel, Oct 12 2019

A110597 Balanced numbers (A020492) k such that k mod 12 = 1.

Original entry on oeis.org

1, 1045, 29029, 50065, 64285, 87685, 1390753, 2011009, 3189625, 7711405, 39298441, 53238625, 68393065, 75416341, 96345613, 225938245, 228404605, 231562825, 233591605, 279999445, 458406445, 462027565, 470527057, 491291125, 513574369, 663605761, 666373825

Offset: 1

Walter Kehowski, Sep 13 2005

nonn

For the first 27 terms, the quotient sigma(n)/phi(n) is 1, 2 or 3.

Amiram Eldar, Table of n, a(n) for n = 1..8224 (terms below 6.5*10^14, calculated using data from Jud McCranie)

Intersection of A017533 and A020492.

Cf. A000010, A000203, A062699.

with(numtheory); BNM1:=[]: for z from 1 to 1 do for m from 0 to 500000 do n:=12*m+1; if sigma(n) mod phi(n) = 0 then BNM1:=[op(BNM1),n] fi; od; od; BNM1;

Select[Range[10^7], Mod[#, 12] == 1 && Divisible[DivisorSigma[1, #], EulerPhi[#]] &] (* Amiram Eldar, Dec 04 2019 *)

forstep(n=1,1e5,12, if(sigma(n)%eulerphi(n)==0, print1(n", "))) \\ Charles R Greathouse IV, Nov 27 2013

a(10)-a(27) from Donovan Johnson, Aug 30 2012

A160080 Lodumo_4 of Fibonacci numbers.

Original entry on oeis.org

0, 1, 5, 2, 3, 9, 4, 13, 17, 6, 7, 21, 8, 25, 29, 10, 11, 33, 12, 37, 41, 14, 15, 45, 16, 49, 53, 18, 19, 57, 20, 61, 65, 22, 23, 69, 24, 73, 77, 26, 27, 81, 28, 85, 89, 30, 31, 93, 32, 97, 101, 34, 35, 105, 36, 109, 113, 38, 39, 117, 40, 121, 125, 42, 43, 129, 44, 133, 137, 46

Offset: 0

Philippe Deléham, May 01 2009

Permutation of nonnegative numbers.

OEIS wiki, Lodumo transform
Index entries for sequences that are permutations of the nonnegative integers
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Cf. A004767, A008586, A016825, A017533, A017581, A017629, A079343.

a(n) = lod_4(A000045(n)).
From Philippe Deléham, Nov 30 2023: (Start)
a(n) = 2*a(n-6) - a(n-12) for n >= 12.
a(6*n) = 4*n, a(6*n+1) = 12*n+1, a(6*n+2) = 12*n+5, a(6*n+3) = 4*n+2, a(6*n+4) = 4*n+3, a(6*n+5) = 12*n+9.
G.f.: (x + 5*x^2 + 2*x^3 + 3*x^4 + 9*x^5 + 4*x^6 + 11*x^7 + 7*x^8 + 2*x^9 + x^10 + 3*x^11) / ((1-x)^2*(1+x+x^2)^2*(1+x^3)^2). (End)

A244923 Numbers n such that the digit sum of Fibonacci(n) is equal to the digit sum of Lucas(n).

Original entry on oeis.org

1, 13, 61, 73, 97, 217, 349, 649, 937, 1477, 1513, 1729, 2005, 2077, 2209, 3265, 3649, 3889, 4093, 4609, 4945, 5497, 5749, 5929, 6109, 7309, 7441, 8041, 8389, 8821, 9925, 10525, 10669, 11605, 13201, 13345, 16021, 18529, 18649, 20293, 21481, 22573, 22729, 24169

Offset: 1

Michel Lagneau, Jul 08 2014

Numbers n such that A004090(n) = A139374(n).
Subsequence of A017533.
It seems that n is odd. The primes of the sequence are: 13, 61, 73, 97, 349, 937, 3889, 4093, 5749, 7309, 8389, 8821, 21481, 22573, 24169, ...
Fibonacci(j) == Lucas(j) (mod 9) iff j == 1 (mod 12), so all a(n) == 1 (mod 12). - Robert Israel, Jul 10 2014

			13 is in the sequence because Fibonacci(13) = 233, Lucas(13) = 521 and 2+3+3 = 5+2+1 = 8.

Vincenzo Librandi, Table of n, a(n) for n = 1..95

Cf. A000032, A000045, A004090, A017533, A139374.

lst={}; Table[If[Total[IntegerDigits[LucasL[n]]] == Total[IntegerDigits[Fibonacci[n]]], AppendTo[lst, n]], {n, 0, 25000}]; lst
Select[Range[25000],Total[IntegerDigits[Fibonacci[#]]]==Total[IntegerDigits[LucasL[#]]]&] (* Harvey P. Dale, Mar 31 2024 *)

A294031 Numbers k such that k == 1 (mod 12) and 6k+1, 12k+1, 18k+1, 36k+1, 72k+1, 108k+1 and 144k+1 are all primes, so N = (6k+1)(12k+1)(18k+1), (36k+1)N, (72k+1)N, (108k+1)N and (144k+1)N are 5 Carmichael numbers in an arithmetic progression.

Original entry on oeis.org

20543425, 80993605, 112608685, 255063865, 307510105, 367621765, 382017685, 400463665, 409631425, 430786405, 536835565, 675787105, 950572525, 1040986765, 1139137825, 1214553025, 1404069205, 1456119805, 1560636805, 1608308905, 1796972905, 1805035225, 1823195605

Offset: 1

Amiram Eldar, Oct 22 2017

nonn

			20543425 generates 11236306070625187487140801 + 8309959597401596721108558352203300 k which are Carmichael numbers for k = 0 to 4.

Andrzej Rotkiewicz, Pseudoprime Numbers and Their Generalizations, Student Association of the Faculty of Sciences, University of Novi Sad, Novi Sad, Yugoslavia, 1972.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Andrzej Rotkiewicz, Arithmetic progressions formed by pseudoprimes, Acta Mathematica et Informatica Universitatis Ostraviensis, Vol. 8, No. 1 (2000), pp. 61-74.

Cf. A002997.

Subsequence of A017533.

aQ[n_]:=Mod[n,12]==1 && AllTrue[{6n+1, 12n+1, 18n+1, 36n+1, 72n+1, 108n+1, 144n+1}, PrimeQ]; Select[Range[10^8], aQ]

A296180 Triangle read by rows: T(n, k) = 3(n - k)k + 1, n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 7, 7, 1, 1, 10, 13, 10, 1, 1, 13, 19, 19, 13, 1, 1, 16, 25, 28, 25, 16, 1, 1, 19, 31, 37, 37, 31, 19, 1, 1, 22, 37, 46, 49, 46, 37, 22, 1, 1, 25, 43, 55, 61, 61, 55, 43, 25, 1, 1, 28, 49, 64, 73, 76, 73, 64, 49, 28, 1

Offset: 0

Wolfdieter Lang, Dec 20 2017

This is member m = 3 of the family of triangles T(m; n, k) = m*(n - k)*k + 1, for m >= 0. For m = 0: A000012(n, k) (read as a triangle); for m = 1: A077028 (rascal), for m = 2: T(2, n+1, k+1) = A130154(n, k). Motivated by A130154 to look at this family of triangles.
In general the recurrence is: T(m; n, 0) = 1 and T(m; n, n) = 1 for n >= 0; T(m; n, k) = (T(m; n-1, k-1)*T(m; n-1, k) + m)/T(m; n-2, k-1), for n >= 2, k = 1..n-1.
The general g.f. of the sequence of column k (with leading zeros) is G(m; k, x) = (x^k/(1 - x)^2)*(1 + (m*k - 1)*x), k >= 0.
The general g.f. of the triangle T(m;, n, k) is GT(m; x, t) = (1 - (1 + t)*x + (m+1)*t*x^2)/((1 - t*x)*(1 - x))^2, and G(m; k, x) = (d/dt)^k GT(m; x, t)/k!|_{t=0}.
For a simple combinatorial interpretation see the one given in A130154 by Rogério Serôdio which can be generalized to m >= 3.

			The triangle T(n, k) begins:
n\k   0  1  2  3  4  5  6  7  8  9 10 ...
0:    1
1:    1  1
2:    1  4  1
3:    1  7  7  1
4:    1 10 13 10  1
5:    1 13 19 19 13  1
6:    1 16 25 28 25 16  1
7:    1 19 31 37 37 31 19  1
8:    1 22 37 46 49 46 37 22  1
9:    1 25 43 55 61 61 55 43 25  1
10:   1 28 49 64 73 76 73 64 49 28  1
...
Recurrence: 28 = T(6, 3) = (19*19 + 3)/13 = 28.

Cf. A077028, A130154.

Columns (without leading zeros): A000012, A016777, A016921, A016921, A017173, A017533, ...

Table[3 k (n - k) + 1, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 20 2017 *)

lista(nn) = for(n=0, nn, for(k=0, n, print1(3*(n - k)*k + 1, ", "))) \\ Iain Fox, Dec 21 2017

T(n, k) = 3*(n - k)*k + 1, n >= 0, 0 <= k <= n,
Recurrence: T(n, 0) = 1 and T(n, n) = 1 for n >= 0; T(n, k) = (T(n-1, k-1)*T(n-1, k) + 3)/T(n-2, k-1), for n >= 2, k = 1..n-1.
G.f. of column k (with leading zeros): (x^k/(1 - x)^2)*(1 + (3*k-1)*x), k >= 0.
G.f. of triangle: (1 - (1 + t)*x + 4*t*x^2)/((1 - t*x)*(1 - x))^2 = 1 + (1+t)*x +(1 + 4*t + t^2)*x^2 + (1 + 7*t + 7*t^2 + t^3)*x^3 = ...

A387404 Numbers of the form 12*k + 1 that satisfy Euler's condition for odd perfect numbers (A228058).

Original entry on oeis.org

325, 637, 925, 1525, 1573, 1813, 1825, 2425, 2725, 2989, 3577, 3757, 3925, 4477, 4525, 4693, 4753, 4825, 5341, 5725, 6025, 6253, 6877, 6925, 7381, 7693, 7825, 8125, 8425, 8725, 8833, 8869, 9325, 9457, 9925, 10225, 10309, 10525, 10693, 10825, 10933, 11221, 11425, 11737, 11809, 12337, 12493, 13189, 13357, 13525, 13573

Offset: 1

Antti Karttunen, Aug 29 2025

Antti Karttunen, Table of n, a(n) for n = 1..10000

Intersection of A017533 and A228058.

nn = 51; n = 1; t = {}; While[Length[t] < nn, n = n + 2; {p, e} = Transpose[FactorInteger[n]]; od = Select[e, OddQ]; If[Length[e] > 1 && Length[od] == 1 && Mod[od[[1]], 4] == 1 && Mod[p[[Position[e, od[[1]]][[1, 1]]]], 4] == 1&&Mod[n,12]==1, AppendTo[t, n]]]; t (* James C. McMahon, Aug 29 2025 *)

is_A387404(n) = if(1!=(n%12) || (omega(n)<2), 0, my(f=factor(n), y=0); for(i=1, #f~, if(1==(f[i, 2]%4), if((1==y)||(1!=(f[i, 1]%4)), return(0), y=1), if(f[i, 2]%2, return(0)))); (y));

cp's OEIS Frontend

A017535 a(n) = (12*n+1)^3.

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A017540 a(n) = (12*n + 1)^8.

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A017543 a(n) = (12*n + 1)^11.

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A033576 a(n) = (2*n+1)*(12*n+1).

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A110597 Balanced numbers (A020492) k such that k mod 12 = 1.

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Maple

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A160080 Lodumo_4 of Fibonacci numbers.

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A244923 Numbers n such that the digit sum of Fibonacci(n) is equal to the digit sum of Lucas(n).

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A294031 Numbers k such that k == 1 (mod 12) and 6*k+1, 12*k+1, 18*k+1, 36*k+1, 72*k+1, 108*k+1 and 144*k+1 are all primes, so N = (6*k+1)*(12*k+1)*(18*k+1), (36*k+1)*N, (72*k+1)*N, (108*k+1)*N and (144*k+1)*N are 5 Carmichael numbers in an arithmetic progression.

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A033576 a(n) = (2n+1)(12*n+1).

A294031 Numbers k such that k == 1 (mod 12) and 6k+1, 12k+1, 18k+1, 36k+1, 72k+1, 108k+1 and 144k+1 are all primes, so N = (6k+1)(12k+1)(18k+1), (36k+1)N, (72k+1)N, (108k+1)N and (144k+1)N are 5 Carmichael numbers in an arithmetic progression.

A296180 Triangle read by rows: T(n, k) = 3(n - k)k + 1, n >= 0, 0 <= k <= n.