A135576 -id:A135576 - cp's OEIS frontend

A001576 a(n) = 1^n + 2^n + 4^n.

3, 7, 21, 73, 273, 1057, 4161, 16513, 65793, 262657, 1049601, 4196353, 16781313, 67117057, 268451841, 1073774593, 4295032833, 17180000257, 68719738881, 274878431233, 1099512676353, 4398048608257, 17592190238721, 70368752566273, 281474993487873, 1125899940397057

Offset: 0

N. J. A. Sloane

Equals A135576, except for the first term. - Omar E. Pol, Nov 18 2008
Conjecture: For n > 1, if a(n) = 1^n + 2^n + 4^n is a prime number then n is of the form 3^h. For example, for h=1, n=3, a(n) = 1^3 + 2^3 + 4^3 = 73 (prime); for h=2, n=9, a(n) = 1^9 + 2^9 + 4^9 = 262657 (prime); for h=3, n=27, a(n) is not prime. - Vincenzo Librandi, Aug 03 2010
The previous conjecture was proved by Golomb in 1978. See A051154. - T. D. Noe, Aug 15 2010
Another more elementary proof can be found in Liu link. - Bernard Schott, Mar 08 2019
Fills in one quarter section of the figurate form of the Sierpinski square curve. See illustration in links and A141725. - John Elias, Mar 29 2023

T. D. Noe, Table of n, a(n) for n = 0..200
John Elias, Illustration of Initial Terms: 1/4 Sierpinski Square Curve
Andy Liu, West German Mathematical Olympiad 1982 - Second round, Problem 4, Crux Mathematicorum, p. 105, Vol. 12, May. 86.
Index entries for linear recurrences with constant coefficients, signature (7,-14,8).

Subsequence of A002061.
Cf. A001550, A034513, A001579, A074501-A074580, A135576, A135577.
See also comments in A051154.
Cf. A006095, A022166.

Mathematica
```
Table[1^n + 2^n + 4^n, {n, 0, 24}]
```

a(n)=1+2^n+4^n \\ Charles R Greathouse IV, Jun 10 2011

[sigma(4,n)for n in range(0,23)] # Zerinvary Lajos, Jun 04 2009

a(n) = 6*a(n-1) - 8*a(n-2) + 3.
O.g.f.: -1/(-1+x) - 1/(-1+2*x) - 1/(-1+4*x) = ( -3+14*x-14*x^2 ) / ( (x-1)*(2*x-1)*(4*x-1) ). - R. J. Mathar, Feb 29 2008
E.g.f.: e^x + e^(2*x) + e^(4*x). - Mohammad K. Azarian, Dec 26 2008
a(n) = A024088(n)/A000225(n). - Reinhard Zumkeller, Feb 15 2009
Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 35*x^2 + 155*x^3 + ... is the o.g.f. for the 2nd subdiagonal of triangle A022166, essentially A006095. - Peter Bala, Apr 07 2015

A135577 Numbers that have only the digit "1" as first, central and final digit. For numbers with 5 or more digits the rest of digits are "0".

Original entry on oeis.org

1, 111, 10101, 1001001, 100010001, 10000100001, 1000001000001, 100000010000001, 10000000100000001, 1000000001000000001, 100000000010000000001, 10000000000100000000001, 1000000000001000000000001, 100000000000010000000000001, 10000000000000100000000000001

Offset: 1

Omar E. Pol, Feb 24 2008

Also, equal to A135576(n), written in base 2.
Essentially the same as A066138. - R. J. Mathar Apr 29 2008
a(n) has 2n-1 digits.

			----------------------------
n ............ a(n)
----------------------------
1 ............. 1
2 ............ 111
3 ........... 10101
4 .......... 1001001
5 ......... 100010001
6 ........ 10000100001
7 ....... 1000001000001
8 ...... 100000010000001
9 ..... 10000000100000001
10 ... 1000000001000000001

G. C. Greubel, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A001576, A066138, A135576, A138118, A138119, A138120, A138147, A138826.

Join[{1}, LinearRecurrence[{111, -1110, 1000}, {111, 10101, 1001001}, 25]] (* G. C. Greubel, Oct 19 2016 *)
Join[{1},Table[FromDigits[Join[{1},PadRight[{},n,0],{1},PadRight[{},n,0],{1}]],{n,0,10}]] (* Harvey P. Dale, Aug 15 2022 *)

Vec(-x*(2000*x^3-1110*x^2+1)/((x-1)*(10*x-1)*(100*x-1))  + O(x^100)) \\ Colin Barker, Sep 16 2013

a(n) = A135576(n), written in base 2.
Also, a(1)=1, for n>1; a(n)=(concatenation of 1, n-2 digits 0, 1, n-2 digits 0 and 1).
From Colin Barker, Sep 16 2013: (Start)
a(n) = 1 + 10^(n-1) + 100^(n-1) for n>1.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>4.
G.f.: x*(2000*x^3 - 1110*x^2 + 1)/((1-x)*(10*x-1)*(100*x-1)). (End)
E.g.f.: (-111 - 200*x + 100*exp(x) + 10*exp(10*x) + exp(100*x))/100. - Elmo R. Oliveira, Jun 13 2025

A144864 a(n) = (4*16^(n-1)-1)/3.

Original entry on oeis.org

1, 21, 341, 5461, 87381, 1398101, 22369621, 357913941, 5726623061, 91625968981, 1466015503701, 23456248059221, 375299968947541, 6004799503160661, 96076792050570581, 1537228672809129301, 24595658764946068821, 393530540239137101141, 6296488643826193618261, 100743818301219097892181

Offset: 1

Artur Jasinski, Sep 23 2008

Old name was: A144863, read as binary numbers, converted to base 10.
All numbers in this sequence for n>1 are congruent to 5 mod 16. - Artur Jasinski, Sep 25 2008
From Omar E. Pol, Sep 10 2011: (Start)
It appears that this is a bisection of A002450.
It appears that this is a bisection of A084241.
It appears that this is a bisection of A153497.
It appears that this is a bisection of A088556, if n>=2.
(End)
All of the above is trivially true. - Joerg Arndt, Aug 19 2014
The aerated sequence (b(n))n>=1 = [1, 0, 21, 0, 341, 0, 5461, 0, 87381, ...] is a fourth-order linear divisibility sequence; that is, a(n) divides a(m) whenever n divides m. It is the case P1 = 0, P2 = -9, Q = -4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Aug 26 2022

Vincenzo Librandi, Table of n, a(n) for n = 1..500
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A001025, A002450, A013776, A056830, A084241, A088556, A094028, A098704, A131865, A135576, A144863, A153497.

Third quadrisection of Jacobsthal numbers A001045; the other quadrisections are A195156 (first), A139792 (second), and A141060 (fourth).

[16^n/12-1/3: n in [1..20]]; // Vincenzo Librandi, Aug 03 2011

Table[1/3 (-1 + 16^(n - 1)) + 16^(n - 1), {n, 1, 17}] (* Artur Jasinski, Sep 25 2008 *)
LinearRecurrence[{17,-16},{1,21},20] (* Harvey P. Dale, Jun 29 2022 *)

vector(66,n,(4*16^(n-1)-1)/3) \\ Joerg Arndt, Aug 19 2014

a(n) = 16^n/12 - 1/3; a(n) = 16*a(n-1) + 5, a(1)=1. - Artur Jasinski, Sep 25 2008
G.f.: x*(1+4*x) / ( (16*x-1)*(x-1) ). - R. J. Mathar, Jan 06 2011
a(n)=b such that Integral_{x=-Pi/2..Pi/2} (-1)^(n+1)*2^(2*n-3)*(cos((2*n-1)*x))/(5/4+sin(x)) dx = c+b*log(3). - Francesco Daddi, Aug 02 2011
a(n) = (2^(4*n-2)-1)/3. - Klaus Purath, Jan 31 2021
From Jianing Song, Aug 30 2022: (Start)
a(n) = A001045(4*n-2).
a(n+1) - a(n) = 10*A013776(n-1) = 20*A001025(n-1) for n >= 1.
a(n) = 10*A098704(n) + 1 = 20*A131865(n-2) + 1 for n >= 2. (End)
E.g.f.: (exp(16*x) - 4*exp(x) + 3)/12. - Stefano Spezia, Apr 18 2024

New name from Joerg Arndt, Aug 19 2014

A153497 a(n) is the number whose binary expansion is A153498(n).

Original entry on oeis.org

1, 7, 21, 73, 341, 1453, 5461, 21157, 87381, 354997, 1398101, 5548693, 22369621, 89828053, 357913941, 1428859477, 5726623061, 22928862037, 91625968981, 366324918613, 1466015503701, 5865493671253, 23456248059221, 93813538989397

Offset: 1

Omar E. Pol, Dec 27 2008

G. C. Greubel, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (5, -14, 50, -56, 80, -64).

Cf. A135576, A153498, A153499, A153500.

LinearRecurrence[{5,-14,50,-56,80,-64},{1,7,21,73,341,1453},30] (* Harvey P. Dale, Nov 21 2012 *)

From R. J. Mathar, Feb 20 2009: (Start)
a(n) = 5*a(n-1) - 14*a(n-2) + 50*a(n-3) - 56*a(n-4) + 80*a(n-5) - 64*a(n-6).
G.f.: x*(1 +2x +16x^3 -24x^4 +32x^5)/((1-x)*(1-4x)*(1+2x^2)*(1+8x^2)). (End)

More terms from R. J. Mathar, Feb 20 2009

A153499 a(n) is the number whose binary expansion is A153500(n).

Original entry on oeis.org

1, 5, 17, 85, 365, 1365, 5285, 21845, 88757, 349525, 1387157, 5592405, 22457045, 89478485, 357214805, 1431655765, 5732215637, 22906492245, 91581229397, 366503875925, 1466373418325, 5864062014805, 23453384746325, 93824992236885

Offset: 1

Omar E. Pol, Dec 27 2008

Empirical g.f. confirmed similar to A153497 and A153498. - Ray Chandler, Oct 15 2024

Index entries for linear recurrences with constant coefficients, signature (5, -14, 50, -56, 80, -64).

Cf. A135576, A153497, A153498, A153500.

Empirical g.f.: -x*(64*x^6-80*x^5+16*x^4-20*x^3-6*x^2-1) / ((x-1)*(4*x-1)*(2*x^2+1)*(8*x^2+1)). - Colin Barker, Sep 17 2013

More terms from R. J. Mathar, Feb 20 2009

A144863 Start with 1, then at each step prepend 10 and append 01.

Original entry on oeis.org

1, 10101, 101010101, 1010101010101, 10101010101010101, 101010101010101010101, 1010101010101010101010101, 10101010101010101010101010101, 101010101010101010101010101010101

Offset: 1

Artur Jasinski, Sep 23 2008, Sep 25 2008

Bisection of A094028. - Omar E. Pol, Nov 12 2008
a(n) is also A144864(n) written in base 2. - Omar E. Pol, Nov 13 2008
Quadrisection of A147759. - Omar E. Pol, Nov 16 2008

Index entries for linear recurrences with constant coefficients, signature (10001,-10000).

Cf. A056830, A094028, A135576, A144864, A147759.

a = {}; k = {1}; Do[x = FromDigits[k, 2]; AppendTo[a, FromDigits[RealDigits[x, 2]]]; AppendTo[k, 0]; AppendTo[k, 1]; PrependTo[k, 0]; PrependTo[k, 1], {n, 1, 100}];
Table[FromDigits[RealDigits[1/12 (-4 + 16^n), 2]], {n, 1, 10}]
a = {}; k = 1; Do[AppendTo[a, k]; k = 10000 k + 101, {n, 1, 10}]; a
Table[1/99 (-1 + 100^(-1 + 2 n)), {n, 1, 20}]
LinearRecurrence[{10001,-10000},{1,10101},20] (* Harvey P. Dale, Aug 22 2014 *)

a(n) = (-1+100^(-1+2*n))/99.
If a(n) is interpreted as binary number, (-4+16^n)/12 gives the decimal representation of a(n).
a(n) = 10000*a(n-1)+101, n>1.
G.f.: x*(1+100*x) / ( (10000*x-1)*(x-1) ).

A376227 a(n) = Product_{k=1..n} (8^k - 1)/(2^k - 1) for n >= 1 with a(0) = 1.

Original entry on oeis.org

1, 7, 147, 10731, 2929563, 3096548091, 12884736606651, 212765655585627963, 13998490777945220569659, 3676801592262757799164923963, 3859174628040582848761303356488763, 16194459027901983959148041623911690081339, 271764285812898926139442499827890355613945218107

Offset: 0

Paul D. Hanna, Oct 09 2024

nonn

			G.f.: A(x) = 1 + 7*x + 147*x^2 + 10731*x^3 + 2929563*x^4 + 3096548091*x^5 + 12884736606651*x^6 + 212765655585627963*x^7 + ...
where the coefficients a(n) of x^n begin
a(0) = 1,
a(1) = 1 * 7,
a(2) = 1 * 7 * 21,
a(3) = 1 * 7 * 21 * 73,
a(4) = 1 * 7 * 21 * 73 * 273,
a(5) = 1 * 7 * 21 * 73 * 273 * 1057,
...

Cf. A376530, A001576, A135576.

{a(n) = (1/3) * prod(k=0,n, 1 + 2^k + 2^(2*k))}
for(n=0,12,print1(a(n),", "))

G.f. A(x) = 1/(1 - 7*x/(1 + 7*x - 21*x/(1 + 21*x - 73*x/(1 + 73*x - 273*x/(1 + 273*x - 1057*x/(1 + 1057*x - 4161*x/(1 + ...))))))), a continued fraction.
a(n) = Product_{k=1..n} (1 + 2^k + 2^(2*k)) for n >= 1 with a(0) = 1.
a(n) = 2^(n*(n+1)/2) * Product_{k=1..n} (1/2^k + 1 + 2^k) for n >= 1.
a(n) ~ c * 2^(n*(n+1)) where c = Product_{n>=1} (1 + 1/2^n + 1/4^n) = 2.975905201850451176749639540825805061981174...

cp's OEIS Frontend

A001576 a(n) = 1^n + 2^n + 4^n.

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A135577 Numbers that have only the digit "1" as first, central and final digit. For numbers with 5 or more digits the rest of digits are "0".

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A144864 a(n) = (4*16^(n-1)-1)/3.

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A153497 a(n) is the number whose binary expansion is A153498(n).

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A153499 a(n) is the number whose binary expansion is A153500(n).

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A144863 Start with 1, then at each step prepend 10 and append 01.

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A376227 a(n) = Product_{k=1..n} (8^k - 1)/(2^k - 1) for n >= 1 with a(0) = 1.

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