Peter Woodward - cp's OEIS frontend

A367897 a(1)=1, a(n) = H_(n-2)(a(n-2), a(n-1)) where H_n is the n-th hyperoperator.

1, 2, 3, 6, 729

Offset: 1

Peter Woodward, Dec 04 2023

nonn

The sequence follows the hierarchy of arithmetic operations (successorship, addition, multiplication, exponentiation, tetration, ...) with the recurrence a(n-2)?a(n-1)=a(n), where "?" follows the sequence of S (successor), +, *, ^, ^^ (tetration), ...
a(6) (=6^^729) is too large to be represented.
Essentially a Fibonacci generalization: the Fibonacci hyperoperation sequence starting with 1.

			a(1)=1, a(2)=S(a(1)), a(3)=a(1)+a(2), a(4)=a(2)*a(3), a(5)=a(3)^a(4), a(6)=a(4)^^a(5), ...a(1) = 1
a(2) = H_0(a(1), a(1)) = 1 + 1 = 2 (successor of 1 = 2)
a(3) = H_1(a(1), a(2)) = 1 + 2 = 3
a(4) = H_2(a(2), a(3)) = 2 * 3 = 6
a(5) = H_3(a(3), a(4)) = 3^6 = 729

Wikipedia, Binary operation
Wikipedia, Hyperoperation
Wikipedia, Generalizations of Fibonacci numbers

Cf. A054871.

A364843 Integers are repeated in runs of 1, 2, 3, ... Each new integer (following a run) is given the value of its sequence index value.

Original entry on oeis.org

1, 2, 2, 4, 4, 4, 7, 7, 7, 7, 11, 11, 11, 11, 11, 16, 16, 16, 16, 16, 16, 22, 22, 22, 22, 22, 22, 22, 29, 29, 29, 29, 29, 29, 29, 29, 37, 37, 37, 37, 37, 37, 37, 37, 37, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56

Offset: 1

Peter Woodward, Aug 10 2023

Omitting repeats yields the triangular numbers plus 1 sequence A000124.

			Illustrated as a triangle begins:
   1;
   2,  2;
   4,  4,  4;
   7,  7,  7,  7;
  11, 11, 11, 11, 11;
  16, 16, 16, 16, 16, 16;
  22, 22, 22, 22, 22, 22, 22;
  ...

Cf. A000124, A002024, A006528.

Row sums give A006000(n-1).

T:= (n, k)-> n*(n-1)/2+1:
seq(seq(T(n,k), k=1..n), n=1..11);  # Alois P. Heinz, Aug 31 2023

a(n) = my(t=(sqrtint(8*n-1)-1)\2); t*(t+1)/2+1 \\ Thomas Scheuerle, Aug 10 2023

from math import isqrt
def A364843(n): return ((t:=isqrt((n<<3)-1)-1>>1)*(t+1)>>1)+1 # Chai Wah Wu, Sep 15 2023

G.f.: x*y*(1 + 2*x^4*y^2 - x*(1 + y) - 2*x^3*y*(1 + y) + x^2*(1 + y + y^2))/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Sep 02 2023

Sum_{k=1..n} k = T(n,k) = A006528(n). - Alois P. Heinz, Sep 15 2023

A356464 Number of black keys in each group of black keys on a standard 88-key piano (left to right).

Original entry on oeis.org

1, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3

Offset: 1

Peter Woodward, Aug 08 2022

On a standard piano keyboard, the black keys appear in groups of two and three, with each group separated from adjacent groups by the presence of two white keys that have no black key between them.
The black keys in a group of two are C#/Db and D#/Eb; the black keys in a group of three are F#/Gb, G#/Ab, and A#/Bb.
The A#/Bb key near the left end of the keyboard is a special case; it is the only black key in its group because the white A key to its left is the leftmost key on the keyboard.

			From _Jon E. Schoenfield_, Aug 12 2022: (Start)
In the diagram below, five octaves (i.e., sets of 12 consecutive keys) have been omitted (as represented by the ellipses):
.
    n |  1       2         3       ...     14        15
  ----+---------------------------------------------------------
  a(n)|  1       2         3       ...      2         3
    ______________________________ ... _________________________
      | |/| | |/||/| | |/||/||/| |     | |/||/| | |/||/||/| |  |
      | |/| | |/||/| | |/||/||/| |     | |/||/| | |/||/||/| |  |
      | |/| | |/||/| | |/||/||/| |     | |/||/| | |/||/||/| |  |
      | |_| | |_||_| | |_||_||_| |     | |_||_| | |_||_||_| |  |
      |  |  |  |  |  |  |  |  |  |     |  |  |  |  |  |  |  |  |
      |  |  |  |  |  |  |  |  |  |     |  |  |  |  |  |  |  |  |
      |__|__|__|__|__|__|__|__|__|     |__|__|__|__|__|__|__|__|
       A  B  C  D  E  F  G  A  B   ...  C  D  E  F  G  A  B  C
(End)

Cf. A059620, A060107, A060106, A081031, A081032.

Cf. A329207.

A346261 Lexicographically earliest sequence of decimal words starting with 10 such that each term has Hamming distance at least 2 from all earlier terms.

Original entry on oeis.org

10, 21, 32, 43, 54, 65, 76, 87, 98, 100, 111, 122, 133, 144, 155, 166, 177, 188, 199, 201, 212, 220, 234, 245, 253, 267, 278, 286, 302, 313, 324, 330, 341, 356, 368, 375, 389, 397, 403, 414, 425, 431, 440, 452, 469, 496, 504, 515, 523, 536, 542, 550, 561, 579, 605, 616, 627, 638, 649, 651

Offset: 1

Peter Woodward, Jul 11 2021

			a(1) = 10 by definition.
a(2) = 21 because '2' has not yet appeared in the tens place and '1' has not yet appeared in the ones place.

Michael S. Branicky, Table of n, a(n) for n = 1..10000
J. H. Conway and N. J. A. Sloane, Lexicographic codes: error-correcting codes from game theory, IEEE Transactions on Information Theory, 32:337-348, 1986.

Lexicodes of minimal distance 1,2,3,... over alphabets of size 2: A001477, A001969, A075926, A075928, A075931, A075934, ...; size 3: A001477, A346002, A346003; size 10: A001477, A343444, A333568, A346000, A346001.

Cf. A207063.

def ham(m, n):
    s, t = str(min(m, n))[::-1], str(max(m, n))[::-1]
    return len(t) - len(s) + sum(s[i] != t[i] for i in range(len(s)))
def aupton(terms):
    alst = [10]
    for n in range(2, terms+1):
        an = alst[-1] + 1
        while any(ham(an, aprev) < 2 for aprev in alst[::-1]):  an += 1
        alst.append(an)
    return alst
print(aupton(60)) # Michael S. Branicky, Jul 22 2021

Edited and corrected by N. J. A. Sloane, Jul 20 2021

A309663 Primes that begin a run of consecutive primes whose first differences are nondecreasing.

Original entry on oeis.org

2, 13, 19, 31, 41, 61, 71, 83, 101, 109, 131, 139, 151, 167, 181, 193, 199, 227, 241, 257, 271, 281, 311, 337, 349, 373, 383, 401, 421, 433, 443, 461, 487, 503, 523, 547, 563, 571, 593, 601, 617, 641, 661, 677, 709, 727, 743, 757, 773, 797, 811, 823, 829, 857

Offset: 1

Peter Woodward, Jun 06 2020

nonn

Arrange primes in rows where the value of the increase between consecutive primes cannot shrink.
Conjecture: Average length of each run = e, converging from above.
Heuristic justification: Consider that e = 2 + 1/2! + 1/3! + 1/4! + ... We always have at least two values in a run. The odds of there being a third value = 1/2!. The odds of there being a fourth value = 1/3! (because exactly 3! ways to sort three differences, and only one of these ways is in increasing order). etc... The process is aberrated by the possibility of equal increases. This sequence allows equal increases in the runs, causing the convergence to e to approach from above. However, as the scale increases, these equal increases occur less frequently and their effect approaches zero. The sister sequence, "Primes that begin a run of consecutive primes whose first differences are strictly increasing", disallows equal increases in its rows, thus it approaches e from below. Averaging the runs of the two sequences negates the aberration, giving immediate convergence to e.

			The first run is 2, 3, 5, 7, 11, thus the first value is 2;
The second run is 13, 17, thus the second value is 13;
The third run is 19, 23, 29, thus the third value is 19;
The fourth run is 31, 37, thus the fourth value is 31.

Reddit blog, Found "e" in the primes (maybe), 2019.

Same process as A331544 except this allows equal increases in the runs.

lista(nn) = {my(d=m=2); forprime(p=2, nn, if(p-mJinyuan Wang, Jul 09 2020

More terms from Jinyuan Wang, Jul 09 2020

A331544 Primes that begin a run of consecutive primes whose first differences are strictly increasing.

Original entry on oeis.org

2, 7, 13, 19, 31, 41, 59, 71, 83, 101, 109, 131, 139, 151, 163, 179, 193, 199, 223, 229, 241, 257, 269, 281, 311, 337, 349, 373, 383, 401, 421, 433, 443, 461, 487, 503, 523, 547, 563, 571, 593, 601, 613, 619, 641, 659, 677, 709, 727, 739, 757, 773, 797, 811, 823

Offset: 1

Peter Woodward, Jan 19 2020

nonn

Arrange primes in rows where the value of the increase between consecutive primes must keep increasing.
Conjecture: Average length of each run = e, converging from below.
Comment from Peter Woodward, Apr 22 2023: (Start)
Heuristic justification: Consider that e = 2 + 1/2! + 1/3! + 1/4! + ... We always have at least two values in a run. The probability that there is a third value is 1/2!. The probability that there is a fourth value is 1/3! (because there are exactly 3! ways to sort three differences, and only one of these ways is in increasing order). And so on. The process is distorted by the possibility of equal increases. This sequence disallows equal increases in the runs, causing the convergence to e to approach from below. However, as the scale increases, these equal increases occur less frequently and their effect approaches zero. The sister sequence, "Primes that begin a run of consecutive primes whose first differences are nondecreasing", includes equal increases in its rows, thus it approaches e from above. Averaging the runs of the two sequences should negate the aberration, giving immediate convergence to e. However, while using a set of random numbers arranged in this fashion does immediately converge to e, with primes the convergence is slowed for unknown reasons, and the average run length remains slightly below e with extensive calculation (after 10^7 runs it reaches 2.712 vs. 2.718...).
The probability that a randomly chosen integer n is prime is approximately 1/log(n) (by the prime number theorem). Thus, the expected gap between consecutive primes of size k is approximately (1/log(n)) * (1 - 1/log(n))^k - 1.
Using the above formula, we can calculate the expected length of the k-th set as follows:
E(length of k-th set) = Sum_{n>=m+1} ((1/log(n)) * (1 - 1/log(n))^k - 1) where m is the largest prime less than or equal to the starting prime of the k-th set.
(End)

			The first run is 2, 3, 5, thus the first value is 2;
the second run is 7, 11, thus the second value is 7;
the third run is 13, 17, thus the third value is 13;
the fourth run is 19, 23, 29, thus the fourth value is 19.

Reddit blog, Found "e" in the primes (maybe), 2019.

Cf. A309663.

lista(nn) = {my(m=2, d=0); forprime(p=2, nn, if(p-m>d, d=p-m, d=0; print1(p, ", ")); m=p); } \\ Jinyuan Wang, Jan 21 2020

More terms from Jinyuan Wang, Jan 21 2020

A260517 Numbers equidistant from twin prime pairs that are also equidistant from numbers equidistant from twin prime pairs.

Original entry on oeis.org

51, 105, 144, 165, 234, 255, 276, 630, 1041, 2289, 2325, 2466, 4251, 5460, 9006, 9699, 10380, 10479, 12006, 13701, 14166, 15690, 18090, 19425, 20190, 20295, 21540, 26706, 26796, 32487, 32871, 33684, 33789, 35520, 37455, 38661, 41685, 42771, 46515, 47760

Offset: 1

Peter Woodward, Jul 27 2015

nonn

			165 is a term because it is equidistant from 144 and 186. 144 and 186 are both equidistant from twin primes, according to A074953.

Robert G. Wilson v, Table of n, a(n) for n = 1..1257

Cf. A001097 (Twin primes), A074953 (Numbers equidistant from twin prime pairs).

t = Select[ Prime@ Range@ 5000, PrimeQ[# + 2] &]; d = Differences@ t; (t[[#+1]] + t[[#+2]]& /@ Select[ Range[ Length[d] - 2], d[[#]] == d[[#+2]] &])/2 + 1 (* Robert G. Wilson v, Jul 29 2015 *)

A261038 a(1)=1; for n>1: a(n) = a(n-1)*n if t=0, a(n) = round(a(n-1)/n) if t=1, a(n) = a(n-1)+n if t=2, a(n) = a(n-1)-n if t=3, where t = n mod 4.

Original entry on oeis.org

1, 3, 0, 0, 0, 6, -1, -8, -1, 9, -2, -24, -2, 12, -3, -48, -3, 15, -4, -80, -4, 18, -5, -120, -5, 21, -6, -168, -6, 24, -7, -224, -7, 27, -8, -288, -8, 30, -9, -360, -9, 33, -10, -440, -10, 36, -11, -528, -11, 39, -12, -624, -12, 42, -13, -728, -13, 45, -14

Offset: 1

Peter Woodward, Aug 07 2015

a(4*n+1) = 1,  0,  -1,  -2,  -3, ...
a(4*n+2) = 3,  6,   9,  12,  15, ...
a(4*n+3) = 0, -1,  -2,  -3,  -4, ...
a(4*n+4) = 0, -8, -24, -48, -80, ... = -A033996(n).

			a(1) =                    1.
a(2) =       a(1) + 2  =  3.
a(3) =       a(2) - 3  =  0.
a(4) =       a(3) * 4  =  0.
a(5) = round(a(4) / 5) =  0.
a(6) =       a(5) + 6  =  6.
a(7) =       a(6) - 7  = -1.

Alois P. Heinz, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A033996.

a:= proc(n) option remember; `if`(n=1, 1, (t->
      `if`(t=0, a(n-1)*n, `if`(t=1, round(a(n-1)/n),
      `if`(t=2, a(n-1)+n, a(n-1)-n))))(irem(n, 4)))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 08 2015

nxt[{n_,a_}]:=Module[{t=Mod[n+1,4]},{n+1,Which[t==0,a(n+1), t==1,Round[ a/(n+1)], t==2,a+n+1,t==3,a-n-1]}]; NestList[nxt,{1,1},100][[All,2]] (* or *) LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{1,3,0,0,0,6,-1,-8,-1,9,-2,-24},100] (* Harvey P. Dale, May 25 2018 *)

Vec(-x*(x^10+2*x^8-8*x^7-x^6-3*x^5-3*x^4+3*x+1)/((x-1)^3*(x+1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Aug 10 2015

first(m)=my(v=vector(m),t);v[1]=1;for(i=2,m,t = i%4;if(t==0,v[i]=v[i-1]*i,if(t==1,v[i]=round(v[i-1]/i),if(t==2,v[i]=v[i-1]+i,v[i]=v[i-1]-i ))));v; \\ Anders Hellström, Aug 17 2015

From Colin Barker, Aug 09 2015: (Start)
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
G.f.: -x*(x^10+2*x^8-8*x^7-x^6-3*x^5-3*x^4+3*x+1) / ((x-1)^3*(x+1)^3*(x^2+1)^3).
(End)

More terms from Alois P. Heinz, Aug 08 2015
Edited by Jon E. Schoenfield, Aug 08 2015
Corrected by Harvey P. Dale, May 25 2018

A260548 a(n) = prime(1)^prime(2)^prime(3)^...^prime(n).

Original entry on oeis.org

2, 8, 14134776518227074636666380005943348126619871175004951664972849610340958208

Offset: 1

Peter Woodward, Jul 29 2015

nonn

			a(1) = 2; a(2) = 2^3; a(3) = 2^3^5.

Cf. A140319, A248012.

A258054 Circle of fifths cycle (counterclockwise).

Original entry on oeis.org

1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8

Offset: 1

Peter Woodward, May 17 2015

The twelve notes dividing the octave are numbered 1 through 12 sequentially. This sequence begins at a certain note, travels down a perfect fifth twelve times (seven semitones), and arrives back at the same note. If justly tuned fifths are used, the final note will be flat by the Pythagorean comma (roughly 23.46 cents or about a quarter of a semitone).
Period 12: repeat [1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8]. - Omar E. Pol, May 18 2015
The string [1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8] is also in both A023127 and A054073. - Omar E. Pol, May 19 2015

			For a(3), 1+5+5 = 11 (mod 12).
For a(4), 1+5+5+5 = 4 (mod 12).

OEIS Wiki, The multi-faceted reach of the OEIS: Music
Wikipedia, Circle of fifths
Wikipedia, Pythagorean comma
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,1).
OEIS index entry for music

Cf. A221363 (Pythagorean comma), A257811 (clockwise circle of fifths cycle).

[1+5*(n-1) mod 12: n in [1..80]]; // Vincenzo Librandi, May 19 2015

A258054:=n->1+(5*(n-1) mod 12): seq(A258054(n), n=1..100); # Wesley Ivan Hurt, May 22 2015

PadRight[{}, 100, {1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8}] (* Vincenzo Librandi, May 19 2015 *)
LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1},{1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8},108] (* Ray Chandler, Aug 27 2015 *)

a(n)=1+5*(n-1) \\ Charles R Greathouse IV, May 22 2015

Vec(x*(1 + 6*x + 11*x^2 + 4*x^3 + 9*x^4 + 2*x^5 + 7*x^6 + 12*x^7 + 5*x^8 + 10*x^9 + 3*x^10 + 8*x^11) / (1 - x^12) + O(x^80)) \\ Colin Barker, Nov 15 2019

Periodic with period 12: a(n) = 1 + (5(n-1) mod 12).
From Colin Barker, Nov 15 2019: (Start)
G.f.: x*(1 + 6*x + 11*x^2 + 4*x^3 + 9*x^4 + 2*x^5 + 7*x^6 + 12*x^7 + 5*x^8 + 10*x^9 + 3*x^10 + 8*x^11) / (1 - x^12).
a(n) = a(n-12) for n>12.
(End)

Extended by Ray Chandler, Aug 27 2015

cp's OEIS Frontend

User: Peter Woodward

A367897 a(1)=1, a(n) = H_(n-2)(a(n-2), a(n-1)) where H_n is the n-th hyperoperator.

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A364843 Integers are repeated in runs of 1, 2, 3, ... Each new integer (following a run) is given the value of its sequence index value.

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A356464 Number of black keys in each group of black keys on a standard 88-key piano (left to right).

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A346261 Lexicographically earliest sequence of decimal words starting with 10 such that each term has Hamming distance at least 2 from all earlier terms.

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A309663 Primes that begin a run of consecutive primes whose first differences are nondecreasing.

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A331544 Primes that begin a run of consecutive primes whose first differences are strictly increasing.

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A260517 Numbers equidistant from twin prime pairs that are also equidistant from numbers equidistant from twin prime pairs.

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A261038 a(1)=1; for n>1: a(n) = a(n-1)*n if t=0, a(n) = round(a(n-1)/n) if t=1, a(n) = a(n-1)+n if t=2, a(n) = a(n-1)-n if t=3, where t = n mod 4.

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Maple