A224486 -id:A224486 - cp's OEIS frontend

A362140 Numbers k in A224486 for which the arithmetic derivative k' (A003415) is also in A224486.

2, 5, 6, 9, 14, 18, 29, 33, 41, 53, 54, 65, 69, 89, 113, 134, 141, 158, 173, 198, 209, 221, 233, 249, 278, 281, 293, 326, 329, 338, 393, 473, 506, 509, 545, 581, 593, 614, 629, 641, 653, 713, 729, 749, 761, 809, 846, 905, 950, 953, 965, 986, 1013, 1014, 1026, 1041, 1049

Offset: 1

Marius A. Burtea, May 03 2023

nonn

Sophie Germain primes p that are not Lucasian primes (A103579) are terms because p' = 1 = A224486(1).

			6 = A224486(4) and 6' = 5 = A224486(3), so 6 is a term.
9 = A224486(5) and 9' = 6 = A224486(4), so 9 is a term.
14 = A224486(6) and 14' = 9 = A224486(5), so 14 is a term.

Cf. A003415, A103579, A224486.

czn:=func; f:=func; [n:n in [2..5000]|czn(n) and czn(Floor(f(n)))];

d[0] = d[1] = 0; d[n_] := n*Plus @@ ((Last[#]/First[#]) & /@ FactorInteger[n]); curzonQ[n_] := PowerMod[2, n, 2*n + 1] == 2*n; Select[Range[2, 1050], curzonQ[#] && curzonQ[d[#]] &] (* Amiram Eldar, May 05 2023 *)

A222948 Numbers k such that 3*k+1 divides 3^k+1.

Original entry on oeis.org

0, 1, 9, 3825, 6561, 102465, 188505, 190905, 1001385, 1556985, 3427137, 5153577, 5270625, 5347881, 13658225, 14178969, 20867625, 23828049, 27511185, 29400657, 48533625, 80817009, 83406609, 89556105, 108464265, 123395265, 127558881, 130747689, 133861905

Offset: 1

Jonathan Vos Post, Apr 07 2013

nonn

This is to 3 as A224486 is to 2

Displayed terms complete up to 200*10^6. - Joerg Arndt, Apr 08 2013

			0 is a term because (3^0+1)/(3*0+1) = 2.
1 is a term because (3^1+1)/(3*1+1) = 1.
9 is a term because (3^9+1)/(3*9+1) = 703.

Joerg Arndt, Table of n, a(n) for n = 1..64 (all terms <= 10^9)

Cf. A224486 (k such that 2*k+1 divides 2^k+1).

Cf. A000244, A016777.

for(n=0,10^9,if((3^n+1)%(3*n+1)==0,print1(n,", "))); /* Joerg Arndt, Apr 08 2013 */
/* the following program is significantly faster; it gives terms >=1: */

for(n=0, 10^12, my(m=3*n+1); if( Mod(3,m)^n==Mod(-1,m), print1(n, ", ") ) ); /* Joerg Arndt, Apr 08 2013 */

{n such that (1+A000244(n))/A016777(n) is an integer}.

Terms > 9 from Joerg Arndt, Apr 08 2013

A224499 Numbers k such that if 2k+1 divides 2^k+1 then 2(k+1)+1 divides 2^(k+1)+1.

Original entry on oeis.org

1, 5, 29, 53, 89, 113, 173, 209, 329, 413, 509, 545, 713, 725, 809, 833, 893, 965, 1013, 1133, 1169, 1625, 1649, 1685, 1733, 1769, 1925, 2009, 2045, 2129, 2273, 2393, 2465, 2549, 2825, 2933, 3065, 3149, 3329, 3389, 3413, 3473, 3605, 3653, 3665, 3773

Offset: 1

Jayanta Basu, Apr 08 2013

nonn

First of a pair of consecutive Curzon numbers, where Curzon numbers are given by A224486.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000051, A224486.

CuQ[n_] := IntegerQ[(2^n + 1)/(2*n + 1)]; Select[Range[4000], CuQ[#] && CuQ[# + 1] &]

A247094 Integers of the form (2^k + 1)/(2k + 1).

Original entry on oeis.org

1, 2, 3, 5, 27, 565, 7085, 48771, 1266205, 9099507, 17602325, 128207979, 26494256091, 11147523830125, 84179432287299, 165269711096165, 281629680514649643, 4246732448623781667, 126774939137440139965, 1925041114036033717685, 14833445639443302757131

Offset: 1

Juri-Stepan Gerasimov, Nov 18 2014

nonn

a(A103579(n)) is a subsequence.

Numbers n such that 2n + 1 divides 2^n + 1: 0, 1, 2, 5, 6, 9, 14, 18, 21, 26, 29, 30, 33, 41, 50, 53, ...

			1 is in this sequence because (2^1 + 1)/(2*1 + 1) = 1,
2 is in this sequence because (2^0 + 1)/(2*0 + 1) = 2,
3 is in this sequence because (2^5 + 1)/(2*5 + 1) = 3.

Colin Barker, Table of n, a(n) for n = 1..400

Cf. A081856, A081858, A103579, A224486, A247132.

s=[]; for(k=0, 100, t=(2^k + 1)/(2*k + 1); if(type(t)=="t_INT", s=concat(s, t))); s=vecsort(s,,8) \\ Colin Barker, Nov 18 2014

a(19) corrected by Colin Barker, Nov 18 2014

A381256 Numbers k such that 5*k+1 divides 5^k+1.

Original entry on oeis.org

0, 1, 625, 57057, 7748433, 30850281, 111494625, 393423745, 499088601, 519341361, 1051107705, 1329416385, 1616038425, 2215448001, 2433936225, 2852972265, 3399207273, 4344683849, 4961725281, 5454760185, 5485530369, 6578054145, 6678031745, 7701979761, 7807302825

Offset: 1

René-Louis Clerc, Feb 18 2025

nonn

The numbers are called Curzon numbers by Tattersall (p. 85, exercise 43).

			5*625+1 = 3126 divides 5^625+1.

James J. Tattersall, Elementary Number Theory in Nine Chapters, Second Edition, Cambridge University Press, 2005, p. 85.

Giovanni Resta, Curzon numbers, Numbers Aplenty.

Cf. A224486, A222948, A230076, A381257, A381258.

PARI
```
isok(n) = my(m=5*n+1); Mod(5, m)^n==-1
```

A381257 Numbers k such that 6*k+1 divides 6^k+1.

Original entry on oeis.org

0, 1, 6, 30, 58, 70, 73, 90, 101, 105, 121, 125, 146, 153, 166, 170, 181, 182, 185, 210, 233, 241, 242, 266, 282, 290, 322, 373, 381, 385, 390, 397, 441, 445, 446, 450, 453, 530, 557, 562, 585, 593, 601, 602, 605, 606, 621, 646, 653, 670, 685, 710, 726, 805, 810, 817, 833, 837, 853, 866

Offset: 1

René-Louis Clerc, Feb 18 2025

nonn

The numbers are called Curzon numbers by Tattersall (p. 85, exercise 43).

			6*30+1 = 181 divides 6^30+1 = 221073919720733357899777.

James J. Tattersall, Elementary Number Theory in Nine Chapters, Second Edition, Cambridge University Press, 2005, p. 85.

Giovanni Resta,Curzon numbers , Numbers Aplenty.

Cf. A224486, A222948, A230076, A381256, A381258.

Select[Range[0, 866], PowerMod[6, #, 6#+1]==6#&]  (* James C. McMahon, Apr 02 2025 *)

PARI
```
isok(n) = my(m=6*n+1); Mod(6, m)^n==-1
```

A381258 Numbers k such that 7*k+1 divides 7^k+1.

Original entry on oeis.org

0, 1, 135, 5733, 11229, 42705, 50445, 117649, 131365, 168093, 636405, 699825, 1269495, 2528155, 4226175, 6176709, 6502545, 9365265, 9551115, 13227021, 14464485, 14912625, 20859435, 26903605, 28251265, 30589905, 32660901, 37597329, 41506875, 42766465, 55452075, 56192535, 111898605

Offset: 1

René-Louis Clerc, Feb 18 2025

nonn

The numbers are called Curzon numbers by Tattersall (p. 85, exercise 43).

James J. Tattersall, Elementary Number Theory in Nine Chapters, Second Edition, Cambridge University Press, 2005, p. 85.

Giovanni Resta, Curzon numbers, Numbers Aplenty.

Cf. A224486, A222948, A230076, A381256, A381257.

Select[Range[0,10^7],PowerMod[7,#,7#+1]==7#&] (* James C. McMahon, Mar 05 2025 *)

PARI
```
isok(n) = my(m=7*n+1); Mod(7, m)^n==-1
```

A385326 The number of positive k <= 2n + 1 such that 2n + 1 divides (2^k + 2*n + 1)^2 - 1.

Original entry on oeis.org

1, 3, 2, 2, 3, 2, 2, 7, 4, 2, 7, 2, 2, 3, 2, 6, 6, 5, 2, 6, 4, 6, 7, 2, 2, 12, 2, 5, 6, 2, 2, 21, 10, 2, 6, 2, 8, 7, 5, 2, 3, 2, 21, 6, 8, 15, 18, 5, 4, 6, 2, 2, 17, 2, 6, 6, 8, 5, 19, 9, 2, 12, 2, 18, 18, 2, 14, 7, 4, 2, 6, 4, 10, 7, 2, 10, 12, 15, 6, 6, 4, 2, 16, 2, 2, 19, 2, 5, 6, 2, 2, 6, 10, 9, 21, 2, 4, 32, 2, 2, 6

Offset: 0

Juri-Stepan Gerasimov, Jun 25 2025

nonn

			1 is the term because 2*0 + 1 = 1 is divisor of (2^1 + 2*0 + 1)^2 - 1 = 3^2 - 1 = 8.

Cf. A003462 (numbers m > 0 such that a(m) = 3), A005384 (primes p such that a(p) = 2), A005408 (odd numbers), A076481 (primes q such that a(q) = 3), A081858 (numbers k numbers k >= 0 such that 2k + 1 divides 2^k - 1), A102781 (numbers k such that 2k + 1 divides (2^k + 2*k + 1)^2 - 1), A224486 (numbers k such that 2k + 1 divides 2^k + 1).

[#[k: k in [1..2*n+1] | ((2^k+2*n+1)^2 - 1) mod (2*n + 1) eq 0]: n in [0..100]];

a[n_]:=Length[Select[Range[2n+1],Divisible[(2^#+2n+1)^2-1,2n+1] &]]; Array[a,101,0] (* Stefano Spezia, Jun 25 2025 *)

a(n) = sum(k=1, 2*n+1, !Mod((2^k + 2*n + 1)^2 - 1, 2*n + 1)); \\ Michel Marcus, Jun 25 2025

cp's OEIS Frontend

A362140 Numbers k in A224486 for which the arithmetic derivative k' (A003415) is also in A224486.

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A222948 Numbers k such that 3*k+1 divides 3^k+1.

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A224499 Numbers k such that if 2*k+1 divides 2^k+1 then 2*(k+1)+1 divides 2^(k+1)+1.

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A247094 Integers of the form (2^k + 1)/(2k + 1).

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A381256 Numbers k such that 5*k+1 divides 5^k+1.

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A381257 Numbers k such that 6*k+1 divides 6^k+1.

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A381258 Numbers k such that 7*k+1 divides 7^k+1.

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A385326 The number of positive k <= 2*n + 1 such that 2*n + 1 divides (2^k + 2*n + 1)^2 - 1.

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A224499 Numbers k such that if 2k+1 divides 2^k+1 then 2(k+1)+1 divides 2^(k+1)+1.

A385326 The number of positive k <= 2n + 1 such that 2n + 1 divides (2^k + 2*n + 1)^2 - 1.