A324298 -id:A324298 - cp's OEIS frontend

A016873 a(n) = 5*n + 2.

2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97, 102, 107, 112, 117, 122, 127, 132, 137, 142, 147, 152, 157, 162, 167, 172, 177, 182, 187, 192, 197, 202, 207, 212, 217, 222, 227, 232, 237, 242, 247, 252, 257, 262, 267, 272, 277

Offset: 0

N. J. A. Sloane

Numbers ending in 2 or 7. - Lekraj Beedassy, Jul 08 2006
For n > 2, also the number of (not necessarily maximal) cliques in the n-gear graph. - Eric W. Weisstein, Nov 29 2017
Also, positive integers k such that 10*k+5 is equal to the product of two integers ending with 5. Proof: if 10*k+5 = (10*a+5) * (10*b+5), then k = 10*a*b + 5*(a+b) + 2 = 5 * (a + b + 2*a*b) + 2, of the form 5m + 2. So, 262 is a term because 2625 = 35 * 75. - Bernard Schott, May 15 2019
Numbers k such that 2^x + 3^x == 0 mod 31 and 2^x + 3^x == 0 mod 11 with x = 6*k+3. - Pedro Caceres, May 18 2022

G. C. Greubel, Table of n, a(n) for n = 0..1000
Cino Hilliard, solutions to 3^x + 5^x == 2 mod 11. [broken link]
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A001622, A008586, A008587, A016861, A342757.
Cf. A053742 (product of two integers ending with 5).
Cf. A324298 (product of two integers ending with 6).

[5*n+2: n in [0..80]]; // G. C. Greubel, Oct 17 2023

a[1]:=2:for n from 2 to 100 do a[n]:=a[n-1]+5 od: seq(a[n], n=1..50); # Zerinvary Lajos, Mar 16 2008

Range[2, 500, 5] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
(* Programs from Eric W. Weisstein, Nov 29 2017 *)
5*Range[0, 70] +2
LinearRecurrence[{2, -1}, {7, 12}, {0, 70}]
CoefficientList[Series[(2+3*x)/(1-x)^2, {x,0,70}], x] (* End *)

a(n)=5*n+2 \\ Charles R Greathouse IV, Jul 10 2016

[i+2 for i in range(300) if gcd(i,5) == 5] # Zerinvary Lajos, May 20 2009

a(n) = 10*n - a(n-1) - 1 (with a(0)=2). - Vincenzo Librandi, Nov 20 2010
G.f.: (2+3*x)/(1-x)^2. - Colin Barker, Jan 08 2012
E.g.f.: exp(x)*(2 + 5*x). - Stefano Spezia, Mar 21 2021
Sum_{n>=0} (-1)^n/a(n) = sqrt(2-2/sqrt(5))*Pi/10 + log(phi)/sqrt(5) - log(2)/5, where phi is the golden ratio (A001622). - Amiram Eldar, Apr 15 2023

A324297 Positive integers k that are the product of two integers ending with 6.

Original entry on oeis.org

36, 96, 156, 216, 256, 276, 336, 396, 416, 456, 516, 576, 636, 676, 696, 736, 756, 816, 876, 896, 936, 996, 1056, 1116, 1176, 1196, 1216, 1236, 1296, 1356, 1376, 1416, 1456, 1476, 1536, 1596, 1656, 1696, 1716, 1776, 1836, 1856, 1896, 1956, 1976, 2016, 2076, 2116

Offset: 1

Stefano Spezia, Mar 16 2019

All the terms end with 6 (A017341).

			36 = 6*6, 96 = 6*16, 216 = 6*36, 256 = 16*16, 276 = 6*46, ...

Stefano Spezia, Table of n, a(n) for n = 1..10000

Cf. A000400, A017341 (supersequence), A324298, A053742 (ending with 5).

a={}; For[n=0,n<=250,n++,For[k=0,k<=n,k++,If[Mod[10*n+6,10*k+6]==0 && Mod[(10*n+6)/(10*k+6),10]==6 && 10*n+6>Max[a],AppendTo[a,10*n+6]]]]; a

isok6(n) = (n%10) == 6; \\ A017341
isok(n) = {if (isok6(n), my(d=divisors(n)); fordiv(n, d, if (isok6(d) && isok6(n/d), return(1)));); return (0);} \\ Michel Marcus, Apr 14 2019

def aupto(lim): return sorted(set(a*b for a in range(6, lim//6+1, 10) for b in range(a, lim//a+1, 10)))
print(aupto(2117)) # Michael S. Branicky, Aug 18 2021

Conjecture: Lim_{n->infinity} a(n)/a(n-1) = 1.

The conjecture is true since it can be proved that a(n) = (sqrt(a(n-1)) + g(n-1))^2 where [g(n): n > 1] is a bounded sequence of positive real numbers. - Stefano Spezia, Aug 18 2021

A346389 a(n) is the number of proper divisors of A324297(n) ending with 6.

Original entry on oeis.org

1, 2, 2, 2, 1, 2, 3, 3, 2, 2, 2, 4, 2, 1, 2, 2, 3, 3, 2, 2, 4, 2, 5, 3, 3, 2, 2, 2, 4, 2, 2, 2, 3, 3, 4, 3, 4, 2, 5, 3, 3, 2, 2, 2, 2, 7, 2, 1, 2, 2, 3, 2, 3, 2, 2, 5, 3, 6, 3, 3, 2, 2, 2, 5, 2, 2, 3, 4, 3, 5, 2, 5, 4, 3, 2, 3, 6, 2, 2, 2, 6, 2, 2, 3, 2, 2, 3, 7

Offset: 1

Stefano Spezia, Jul 15 2021

			a(12) = 4 since there are 4 proper divisors of A324297(12) = 576 ending with 6: 6, 16, 36 and 96.

Cf. A017341, A032741, A324297, A324298, A337856, A346388 (ending with 5), A346392.

b={}; For[n=0, n<=450, n++, For[k=0, k<=n, k++, If[Mod[10*n+6, 10*k+6]==0 && Mod[(10*n+6)/(10*k+6), 10]==6 && 10*n+6>Max[b], AppendTo[b, 10*n+6]]]]; (* A324297 *) a={}; For[i =1, i<=Length[b], i++, AppendTo[a, Length[Drop[Select[Divisors[Part[b, i]], (Mod[#,10]==6&)], -1]]]]; a

a(n) = A346392(A324297(n)).

A346508 Positive integers k such that 10*k+1 is equal to the product of two integers greater than 1 and ending with 1 (A346507).

Original entry on oeis.org

12, 23, 34, 44, 45, 56, 65, 67, 78, 86, 89, 96, 100, 107, 111, 122, 127, 128, 133, 144, 149, 155, 158, 166, 168, 170, 177, 188, 189, 191, 199, 209, 210, 212, 220, 221, 232, 233, 243, 250, 251, 254, 260, 265, 275, 276, 282, 287, 291, 296, 298, 309, 311, 313, 317

Offset: 1

Stefano Spezia, Jul 21 2021

			107 is a term because 21*51 = 1071 = 107*10 + 1.

Stefano Spezia, Table of n, a(n) for n = 1..10000

Cf. A016873 (ending with 5), A017281, A324298 (ending with 6), A346507, A346509, A346510.

a={}; For[n=1, n<=350, n++, For[k=1, kMax[10a+1], AppendTo[a, n]]]]; a

def aupto(lim): return sorted(set(a*b//10 for a in range(11, 10*lim//11+2, 10) for b in range(a, 10*lim//a+2, 10) if a*b//10 <= lim))
print(aupto(318)) # Michael S. Branicky, Aug 21 2021

a(n) = (A346507(n) - 1)/10.
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 1.
The conjecture is true since a(n) = (A346507(n) - 1)/10 and lim_{n->infinity} A346507(n)/A346507(n-1) = 1. - Stefano Spezia, Aug 21 2021

A346951 Positive integers k such that 10*k+9 is equal to the product of two integers ending with 3 (A346950).

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 16, 18, 21, 24, 27, 29, 30, 33, 36, 39, 42, 45, 48, 51, 52, 54, 55, 57, 60, 63, 66, 68, 69, 72, 75, 78, 81, 84, 87, 90, 93, 94, 96, 98, 99, 102, 105, 107, 108, 111, 114, 117, 120, 121, 123, 126, 129, 132, 133, 135, 138, 141, 144, 146, 147, 150

Offset: 1

Stefano Spezia, Aug 08 2021

			15 is a term because 3*53 = 159 = 15*10 + 9.

Cf. A016873 (ending with 5), A017377, A324298 (ending with 6), A346950, A346952, A346953.

a={}; For[n=0, n<=150, n++, For[k=0, k<=n, k++, If[Mod[10*n+9, 10*k+3]==0 && Mod[(10*n+9)/(10*k+3), 10]==3&& 10*n+9>Max[10a+9], AppendTo[a, n]]]]; a

def aupto(lim): return sorted(set(a*b//10 for a in range(3, 10*lim//3+4, 10) for b in range(a, 10*lim//a+4, 10) if a*b//10 <= lim))
print(aupto(150)) # Michael S. Branicky, Aug 11 2021

a(n) = (A346950(n) - 9)/10.

Lim_{n->infinity} a(n)/a(n-1) = 1.

A347254 Positive integers k such that 10*k+6 is equal to the product of two integers ending with 4 (A347253).

Original entry on oeis.org

1, 5, 9, 13, 17, 19, 21, 25, 29, 33, 37, 41, 45, 47, 49, 53, 57, 61, 65, 69, 73, 75, 77, 81, 85, 89, 93, 97, 101, 103, 105, 109, 113, 115, 117, 121, 125, 129, 131, 133, 137, 141, 145, 149, 153, 157, 159, 161, 165, 169, 173, 177, 181, 183, 185, 187, 189, 193, 197

Offset: 1

Stefano Spezia, Aug 24 2021

Since an integer 10*k + 6 = (10*a + 4)*(10*b + 4) implies that k = 10*a*b + 4*(a + b) + 1, all the terms of this sequence are odd.

			13 is a term because 4*34 = 136 = 13*10 + 6.

Cf. A016873 (ending with 5), A017341, A324298 (ending with 6), A346951 (ending with 3), A347253.

a={}; For[n=0, n<=200, n++, For[k=0, k<=n, k++, If[Mod[10*n+6, 10*k+4]==0 && Mod[(10*n+6)/(10*k+4), 10]==4 && n>Max[a], AppendTo[a, n]]]]; a

isok(k) =  my(x=10*k+6); sumdiv(x, d, (Mod(d, 10)==4) && Mod(x/d, 10)==4); \\ Michel Marcus, Oct 04 2021

def aupto(lim): return sorted(set(a*b//10 for a in range(4, 10*lim//4+3, 10) for b in range(a, 10*lim//a+3, 10) if a*b//10 <= lim))
print(aupto(197)) # Michael S. Branicky, Aug 24 2021

a(n) = (A347253(n) - 6)/10.

Lim_{n->infinity} a(n)/a(n-1) = 1.

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A016873 a(n) = 5*n + 2.

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A324297 Positive integers k that are the product of two integers ending with 6.

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A346389 a(n) is the number of proper divisors of A324297(n) ending with 6.

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A346508 Positive integers k such that 10*k+1 is equal to the product of two integers greater than 1 and ending with 1 (A346507).

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A346951 Positive integers k such that 10*k+9 is equal to the product of two integers ending with 3 (A346950).

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A347254 Positive integers k such that 10*k+6 is equal to the product of two integers ending with 4 (A347253).

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