A020714 -id:A020714 - cp's OEIS frontend

A135089 Triangle T(n,k) = 5*binomial(n,k) with T(0,0) = 1, read by rows.

1, 5, 5, 5, 10, 5, 5, 15, 15, 5, 5, 20, 30, 20, 5, 5, 25, 50, 50, 25, 5, 5, 30, 75, 100, 75, 30, 5, 5, 35, 105, 175, 175, 105, 35, 5, 5, 40, 140, 280, 350, 280, 140, 40, 5, 5, 45, 180, 420, 630, 630, 420, 180, 45, 5, 5, 50, 225, 600, 1050, 1260, 1050, 600, 225, 50, 5

Offset: 0

Gary W. Adamson, Nov 18 2007

Row sums = A020714 (except for the first term).

Triangle T(n,k), 0 <= k <= n, read by rows given by (5, -4, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (5, -4, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 24 2013

			First few rows of the triangle:
  1;
  5,  5;
  5, 10,  5;
  5, 15, 15,   5;
  5, 20, 30   20,  5;
  5, 25, 50,  50, 25,  5;
  5, 30, 75, 100, 75, 30, 5.

G. C. Greubel, Table of n, a(n) for the first 50 rows

Cf. A007318, A020714, A132200, A134058, A134059.

[1] cat [5*Binomial(n,k): k in [0..n], n in [1..12]]; // G. C. Greubel, May 03 2021

Table[5*Binomial[n,k] -4*Boole[n==0], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Sep 22 2016; May 03 2021 *)

def A135089(n,k): return 5*binomial(n,k) - 4*bool(n==0)
flatten([[A135089(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 03 2021

T(n,k) = 5*binomial(n,k), n > 0, 0 <= k <= n.
Equals 2*A134059(n,k) - A007318(n,k).
G.f.: (1+4*x+4*x*y)/(1-x-x*y). - Philippe Deléham, Nov 24 2013
Sum_{k=0..n} T(n,k) = A020714(n) - 4*[n=0]. - G. C. Greubel, May 03 2021

A146763 Rank of terms ending in 0 in A061039.

Original entry on oeis.org

0, 4, 10, 14, 20, 24, 30, 34, 40, 44, 50, 54, 60, 64, 70, 74, 80, 84, 90, 94, 100, 104, 110, 114, 120, 124, 130, 134, 140, 144, 150, 154, 160, 164, 170, 174, 180, 184, 190, 194, 200, 204, 210, 214, 220, 224, 230, 234, 240, 244, 250, 254, 260, 264, 270, 274

Offset: 0

Paul Curtz, Nov 02 2008

From Paschen spectrum of hydrogen.

Numbers that are congruent to 0 or 4 mod 10. - Philippe Deléham, Oct 18 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001622, A020714, A030308, A061039.

[5*n - (n mod 2): n in [0..60]]; // G. C. Greubel, Mar 10 2022

Select[Range[0, 100], MemberQ[{0,4}, Mod[#, 10]] &] (* K G Teal, Dec 02 2014 *)

[5*n - (n%2) for n in (0..60)] # G. C. Greubel, Mar 10 2022

a(n) = 10*n - 6 - a(n-1) (with a(0)=0). - Vincenzo Librandi, Nov 26 2010
a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=4 and b(k) = 5*2^k = A020714(k) for k>0. - Philippe Deléham, Oct 18 2011
From Colin Barker, May 14 2012: (Start)
a(n) = (-1 + (-1)^n + 10*n)/2.
a(n) = a(n-1) + a(n-2) - a(n-3).
G.f.: x*(4+6*x)/((1-x)^2*(1+x)). (End)
E.g.f.: 1/2 (exp(-x) - (1 - 10*x)*exp(x)). - G. C. Greubel, Mar 10 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(1-2/sqrt(5))*Pi/20 - log(phi)/(4*sqrt(5)) + log(5)/8, where phi is the golden ratio (A001622). - Amiram Eldar, Sep 17 2023

A166577 Inverse binomial transform of A166517.

Original entry on oeis.org

1, 4, -5, 10, -20, 40, -80, 160, -320, 640, -1280, 2560, -5120, 10240, -20480, 40960, -81920, 163840, -327680, 655360, -1310720, 2621440, -5242880, 10485760, -20971520, 41943040, -83886080, 167772160, -335544320, 671088640, -1342177280, 2684354560, -5368709120

Offset: 0

Paul Curtz, Oct 17 2009

The definition assumes that the offset of A166517 is changed to 0.
A166517 mod 9 yields a periodic sequence with period 1, 5, 4, 8, 7, 2.
This set of numbers in the period appears also in A153130, A146501, and A166304.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-2).

Cf. A010716, A010692, A010859.

Join[{1,4},NestList[-2#&,-5,40]] (* Harvey P. Dale, Aug 02 2012 *)
Join[{1, 4}, LinearRecurrence[{-2}, {-5}, 48]] (* G. C. Greubel, May 17 2016 *)

a(n) = -2*a(n-1), n>2.
a(n) = (-1)^(n+1)*A020714(n-2), n>1.
From Colin Barker, Jan 07 2013: (Start)
a(n) = -5*(-1)^n*2^(n-2) for n>1.
G.f.: (3*x^2+6*x+1)/(2*x+1). (End)
E.g.f.: (9/4) + (3/2)*x - (5/4)*exp(-2*x). - Alejandro J. Becerra Jr., Feb 15 2021

Edited, comments not concerning this sequence removed, and extended by R. J. Mathar, Oct 21 2009

A188381 Negabinary Keith numbers.

Original entry on oeis.org

2, 3, 4, 7, 9, 13, 16, 36, 55, 64, 162, 256, 458, 1024, 1829, 4096, 7316, 15119, 16384, 18970, 37702, 37723, 45171, 60476, 65536, 84506, 262144, 277263, 1048576, 1109052, 1722002, 2160570, 4194304, 10549178, 12699958, 15084573, 16777216, 31921069, 67108864

Offset: 1

Alonso del Arte, Mar 29 2011

Keith numbers are described in A007629. All powers of 4 appear. However, 2 is the only number of the form 2^n with n odd that appears in the sequence. That's because in negabinary, such numbers are represented as 11 followed by n 0's, and that leads to the sequence 1, 1, 0, ... , 0, 2, 3, 5, 10, 20, 40, 80, 160, ... up to 5(2^(n - 2)), and 5(2^(n - 2)) > 2^(n - 1). (See A020714).

Cf. A007629, A039724, A162724.

(* First run the program from A039724 to define ToNegaBases *) keithFromListQ[n_Integer, digits_List] := Module[{seq = digits, curr = digits[[-1]], ord = Length[digits]}, While[curr < n, curr = Plus@@Take[seq, -ord]; AppendTo[seq, curr]]; Return[seq[[-1]] == n]]; Select[Range[2, 32768], keithFromListQ[#, IntegerDigits[ToNegaBases[#, 2]]] &]

a(33)-a(39) from Amiram Eldar, Jan 29 2020

A213948 Triangle, by rows, generated from the INVERT transforms of (1, 1, 2, 4, 8, 16, ...).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 4, 4, 4, 1, 7, 10, 8, 8, 1, 12, 24, 20, 16, 16, 1, 20, 52, 56, 40, 32, 32, 1, 33, 112, 144, 112, 80, 64, 64, 1, 54, 238, 344, 320, 224, 160, 128, 128, 1, 88, 496, 828, 848, 640, 448, 320, 256, 256

Offset: 1

Gary W. Adamson, Jun 25 2012

Row sums = A001519, the odd-indexed Fibonacci terms. The triangle is a companion to A213947, having row sums of the even-indexed Fibonacci terms.

			First few rows of the triangle are:
  1;
  1,   1;
  1,   2,   2;
  1,   4,   4,   4;
  1,   7,  20,   8,   8;
  1,  12,  24,  20,  16,  16;
  1,  20,  52,  56,  40,  32,  32;
  1,  33, 112, 144, 112,  80,  64,  64;
  1,  54, 238, 344, 320, 224, 160, 128, 128;
  1,  88, 496, 828, 848, 640, 448, 320, 256, 256;
  ...

Cf. A001519, A213947, A000071 (2nd column), A020714 (subdiagonal), A005009 (subdiagonal).

read("transforms") ;
A213948i := proc(n,k)
    if n = 1 then
        L := [1,seq(0,i=0..k)] ;
    else
        L := [1,seq(2^i,i=0..n-2),seq(0,i=0..k)] ;
    end if;
    INVERT(L) ;
    op(k,%) ;
end proc:
A213948 := proc(n,k)
    if k = 1 then
        1;
    else
        A213948i(k,n)-A213948i(k-1,n) ;
    end if;
end proc: # R. J. Mathar, Jun 30 2012

Create an array in which the n-th row is the INVERT transform of the first n terms in the sequence (1, 1, 2, 4, 8, 16, 32, ...):
  1,    1,    1,    1,    1,    1,
  1,    2,    3,    5,    8,   13,    (essentially A000045)
  1,    2,    5,    9,   18,   37,    (essentially A077947)
  1,    2,    5,   13,   26,   57,
Terms of the n-th row of the triangle are the finite differences downwards the n-th column of this array.

A287798 Least k such that A006667(k)/A006577(k) = 1/n.

Original entry on oeis.org

159, 6, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 655360, 1310720, 2621440, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 335544320, 671088640, 1342177280, 2684354560, 5368709120, 10737418240

Offset: 3

Michel Lagneau, Jun 01 2017

A006667: number of tripling steps to reach 1 in '3x+1' problem.
A006577: number of halving and tripling steps to reach 1 in '3x+1' problem.
a(n) = {159, 6} union {A020714}.

			a(3) = 159 because A006667(159)/A006577(159) = 18/54 = 1/3.

Colin Barker, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A006577, A006666, A006667. Essentially the same as A020714, A084215, A146523 and A257113.

nn:=10^12:
for n from 3 to 35 do:
ii:=0:
for k from 2 to 10^6 while(ii=0) do:
  m:=k:s1:=0:s2:=0:
   for i from 1 to nn while(m<>1) do:
    if irem(m,2)=0
     then
     s2:=s2+1:m:=m/2:
     else
     s1:=s1+1:m:=3*m+1:
    fi:
   od:
    if n*s1=s1+s2
     then
     ii:=1: printf(`%d, `,k):
     else
    fi:
od:od:

f[u_]:=Module[{a=u,k=0},While[a!=1,k++;If[EvenQ[a],a=a/2,a=a*3+1]];k];Table[f[u],{u,10^7}];g[v_]:=Count[Differences[NestWhileList[If[EvenQ[#],#/2,3#+1]&,v,#>1&]],_?Positive];Table[g[v],{v,10^7}];Do[k=3;While[g[k]/f[k]!=1/n,k++];Print[n," ",k],{n,3,35}]

a(n) = if(n < 5, [0,0,159,6][n], 5<<(n-5)) \\ David A. Corneth, Jun 01 2017

Vec(x^3*(159 - 312*x - 7*x^2) / (1 - 2*x) + O(x^50)) \\ Colin Barker, Jun 01 2017

For n >= 5, a(n) = 5*2^n/32. - David A. Corneth, Jun 01 2017
From Colin Barker, Jun 01 2017: (Start)
G.f.: x^3*(159 - 312*x - 7*x^2) / (1 - 2*x).
a(n) = 2*a(n-1) for n>5.
(End)

A309786 a(n) is the length of the cycle of the trajectory of 1/n under the map f(x) = min(2x, 2-2x).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 1, 3, 2, 5, 1, 6, 3, 4, 1, 4, 3, 9, 2, 6, 5, 11, 1, 10, 6, 9, 3, 14, 4, 5, 1, 5, 4, 12, 3, 18, 9, 12, 2, 10, 6, 7, 5, 12, 11, 23, 1, 21, 10, 8, 6, 26, 9, 20, 3, 9, 14, 29, 4, 30, 5, 6, 1, 6, 5, 33, 4, 22, 12, 35, 3, 9, 18, 20, 9, 30, 12

Offset: 1

Rémy Sigrist, Aug 17 2019

nonn

For any rational number q in the interval [0, 1]:
- f(q) is rational and lies in the interval [0, 1],
- denominator(f(q)) <= denominator(q),
- as there are only finitely many rational numbers whose denominator is less than or equal to that of q in the interval [0, 1],
- iteratively applying f to q eventually leads to a cycle,
- and the sequence is well defined.
For any irrational number x in the interval [0, 1]:
- f(x) is irrational and lies in the interval [0, 1],
- for any k > 0, as fixed points of the k-th iterate of f are rational,
- iteratively applying f to x never leads to a cycle.
As f is continuous on the interval [0, 1] and 1/7 has least period 3, according to the period three theorem, f has points of any period length, as well as chaotic points.

			For n = 5:
- f(1/5) = 2/5,
- f(2/5) = 4/5,
- f(4/5) = 2/5,
- hence a(5) = 2.

Roman Khrabrov, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Period Three Theorem

Cf. A003558, A020714, A029744.

a(n, f=x -> min(2*x, 2-2*x)) = my (x=f(1/n), y=f(x)); while (x!=y, x=f(x); y=f(f(y))); for (k=1, oo, if (x==y=f(y), return (k)))

a(2*n-1) = A003558(n-1).
a(2*n) = a(n).
a(n) = 1 iff n belongs to A029744.
a(n) = 2 iff n belongs to A020714.

A321643 a(n) = 5*2^n - (-1)^n.

Original entry on oeis.org

4, 11, 19, 41, 79, 161, 319, 641, 1279, 2561, 5119, 10241, 20479, 40961, 81919, 163841, 327679, 655361, 1310719, 2621441, 5242879, 10485761, 20971519, 41943041, 83886079, 167772161, 335544319, 671088641, 1342177279, 2684354561, 5368709119, 10737418241, 21474836479

Offset: 0

Paul Curtz, Dec 03 2018

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2).

Cf. A010690, A010701, A020714, A051049, A070366, A110286.

Cf. A001045, A081808, A321483.

List([0..30],n->5*2^n-(-1)^n); # Muniru A Asiru, Dec 05 2018

[5*2^n-(-1)^n$n=0..30]; # Muniru A Asiru, Dec 05 2018

a[n_] := 5*2^n - (-1)^n; Array[a, 30, 0] (* Amiram Eldar, Dec 03 2018 *)

Vec((4 + 7*x) / ((1 + x)*(1 - 2*x)) + O(x^40)) \\ Colin Barker, Dec 04 2018

for n in range(0,30): print(5*2**n - (-1)**n) # Stefano Spezia, Dec 05 2018

a(n+2) - a(n) = a(n+1) + a(n) = 15*2^n, n >= 0.
a(n) - 2*a(n-1) = period 2: repeat [3, -3], n > 0, a(0)=4, a(1)=11.
a(n+1) = 10*A051049(n) + period 2: repeat [1, 9].
a(n) = 12*2^n - A321483(n), n >= 0.
a(n) = 2^(n+2) + 3*A001045(n), n >= 0.
a(n) == A070366(n+4) (mod 9).
From Colin Barker, Dec 04 2018: (Start)
G.f.: (4 + 7*x) / ((1 + x)*(1 - 2*x)).
a(n) = a(n-1) + 2*a(n-2) for n > 1. (End)
E.g.f.: exp(-x)*(5*exp(3*x) - 1). - Elmo R. Oliveira, Aug 17 2024

A335491 a(n) is the smallest number m with exactly n divisors whose last digit equals the last digit of m.

Original entry on oeis.org

1, 11, 40, 60, 160, 120, 640, 240, 360, 480, 8064, 600, 18144, 1920, 1440, 1200, 72576, 1800, 52416, 2400, 5760, 30720, 183456, 3600, 12960, 122880, 9000, 9600, 602784, 7200, 445536, 8400, 92160, 798336, 51840, 12600, 2159136, 576576, 368640, 16800, 2935296, 28800

Offset: 1

Bernard Schott, Jun 11 2020

a(n) exists for any n >= 1. Indeed, the number 5*2^n (see A020714), n >= 1, has exactly n divisors (5*2^1, 5*2^2, ..., 5*2^n), with the last digit 0. - Marius A. Burtea, Jun 12 2020
It's always the case when a(n) ends in 0 then a(n) = 10 * A005179(n). Proof: Let v be the list of divisors of a(n) that end in 0. We then have |v| = n and lcm(v) = a(n) as a(n) is in v and all other terms in v divide a(n). We then have lcm(v)/10 = a(n)/10 where a(n)/10 has exactly n divisors. The least positive integer that has exactly n divisors is A005179(n). - Bernard Schott and David A. Corneth, Jun 12 2020
For some p_i^e_i||a(n) and p_j^e_j||a(n) where p_i and p_j are primes such that p_i < p_j, 10 | (p_j - p_i) and t^k||u denotes t^k|u but t^(k + 1) doesn't divide u i.e. gcd(t, u/t^k) = 1 denotes then e_i >= e_j. For example k * 11^2 * 31^3 where gcd(k, 11*31) = 1 can't be a term as the multiplicity of 11 is less than the multiplicity of 31. - David A. Corneth, Jun 12 2020

			Of the twelve divisors of 60, four have their last digit equals to the last digit of 60: 10, 20, 30 and 60, and there is no smaller number k with four divisors whose last digit equals the last digit of k, hence a(4) = 60.

Project Euler, Problem 474: Last digits of divisors

Cf. A000005, A005179, A020714, A330348.

Similar with: A333456 (Niven numbers), A335038 (Zuckerman numbers).

a:=[]; for n in [1..30] do k:=1; while #[d:d in Divisors(k)|k mod 10 eq d mod 10] ne n do k:=k+1; end while; Append(~a,k); end for; a; // Marius A. Burtea, Jun 12 2020

d[n_] := DivisorSum[n, 1 &, Mod[# - n, 10] == 0 &]; mx = 20; c = 0; n = 1; s = Table[0, {mx}]; While[c < mx, i = d[n]; If[i <= mx && s[[i]] == 0, c++; s[[i]] = n]; n++]; s (* Amiram Eldar, Jun 12 2020 *)

f(n) = my(u=n%10); sumdiv(n, d, (d%10) == u);
a(n) = my(k=1); while(f(k) != n, k++); k; \\ Michel Marcus, Jun 12 2020

Corrected and extended by Marius A. Burtea, Jun 12 2020

A341329 Numbers k such that k^2 is the sum of m nonzero squares for all 1 <= m <= k^2 - 14.

Original entry on oeis.org

13, 15, 17, 25, 26, 29, 30, 34, 35, 37, 39, 41, 45, 50, 51, 52, 53, 55, 58, 60, 61, 65, 68, 70, 73, 74, 75, 78, 82, 85, 87, 89, 90, 91, 95, 97, 100, 101, 102, 104, 105, 106, 109, 110, 111, 113, 115, 116, 117, 119, 120, 122, 123, 125, 130, 135, 136, 137

Offset: 1

Jianing Song, Feb 09 2021

Numbers k such that k^2 is in A018820. Note that k^2 is never the sum of k^2 - 13 positive squares.
A square k^2 is the sum of m positive squares for all 1 <= m <= k^2 - 14 if k^2 is the sum of 2 and 3 positive squares (see A309778 and proof in Kuczma).
Intersection of A009003 and A005767. Also A009003 \ A020714.
Numbers k not of the form 5*2^e such that k has at least one prime factor congruent to 1 modulo 4.
Has density 1 over all positive integers.

			13 is a term: 169 = 13^2 = 5^2 + 12^2 = 3^2 + 4^2 + 12^2 = 11^2 + 4^2 + 4^2 + 4^2 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2 = 6^2 + 6^2 + 6^2 + 6^2 + 4^2 + 3^2 = ... = 3^2 + 2^2 + 2^2 + 1^2 + 1^2 + ... + 1^2 (sum of 155 positive squares, with 152 (1^2)'s), but 169 cannot be represented as the sum of 156 positive squares.

Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 76-79.

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A018820, A309778, A009003, A005767, A020714.

isA341329(n) = setsearch(Set(factor(n)[, 1]%4), 1) && !(n/5 == 2^valuation(n, 2))

cp's OEIS Frontend

A135089 Triangle T(n,k) = 5*binomial(n,k) with T(0,0) = 1, read by rows.

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Sage

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A146763 Rank of terms ending in 0 in A061039.

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Sage

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A166577 Inverse binomial transform of A166517.

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A188381 Negabinary Keith numbers.

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A213948 Triangle, by rows, generated from the INVERT transforms of (1, 1, 2, 4, 8, 16, ...).

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Maple

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A287798 Least k such that A006667(k)/A006577(k) = 1/n.

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Maple

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A309786 a(n) is the length of the cycle of the trajectory of 1/n under the map f(x) = min(2*x, 2-2*x).

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A309786 a(n) is the length of the cycle of the trajectory of 1/n under the map f(x) = min(2x, 2-2x).