Jim Singh - cp's OEIS frontend

A320277 A digitized pure tuning tone, sampled at standard settings for consumer audio: a(n) = floor(sin(2Pi(440/44100)n)32767).

0, 2052, 4097, 6126, 8130, 10103, 12036, 13921, 15752, 17521, 19222, 20846, 22389, 23844, 25205, 26467, 27625, 28675, 29612, 30433, 31134, 31713, 32167, 32494, 32695, 32766, 32709, 32524, 32210, 31770, 31206, 30518, 29711, 28787, 27750, 26604, 25354, 24004, 22559, 21026, 19410

Offset: 0

Jim Singh, Oct 08 2018

sign

This sequence represents sample values for the simplest sound: a pure tone with no harmonics (i.e., a sine wave) with 0 phase shift, of digital peak amplitude, pitched at the standard tuning frequency (A440), and sampled at the standard sampling rate and bit depth resolution for consumer audio: 44100 Hz and 16 bits, respectively.
Since the maximum signed integer value that can be stored in 16 bits is (2^15)-1=32767, the method used to convert the floats to integers is to multiply the floats by 32767 then cast to integer.
The numerator and the denominator of the g.f. have degrees respectively equal to 2204 and 2205. - Stefano Spezia, Nov 02 2018

A. Davison, Killer Game Programming in Java, O'Reilly Media, 2005, page 254.

KVR, How to get each sample of a sine wave given sample rate, frequency, bit depth
Wikipedia, Audio bit depth
Wikipedia, Pulse-code modulation
Wikipedia, Sampling

a[n_]:= Floor[Sin[2*Pi*(440/44100)*n]*32767]; Array[a, 50, 0] (* Stefano Spezia, Nov 02 2018 *)

import math
for x in range(100):
    floatval = math.sin(2*math.pi*(440/44100)*x)*32767
    intval = int(floatval)
    print(intval, end=', ')

A320170 Every pair of consecutive numbers sums to a palindrome, starting with a(0)=1; a(1)=10, and taking a(n) = smallest number > a(n-2).

Original entry on oeis.org

1, 10, 12, 21, 23, 32, 34, 43, 45, 54, 47, 64, 57, 74, 67, 84, 77, 94, 87, 104, 98, 114, 108, 124, 118, 134, 128, 144, 138, 154, 149, 164, 159, 174, 169, 184, 179, 194, 189, 204, 200, 214, 210, 224, 220, 234, 230, 244, 240, 254, 251, 264, 261, 274, 271

Offset: 0

Jim Singh, Oct 07 2018

Cf. A062932.

For n < 10, equals A061870 and A175885.

Nest[Append[#, Block[{k = #[[-2]] + 1}, While[! PalindromeQ[#[[-1]] + k], k++]; k]] &, {1, 10}, 53] (* Michael De Vlieger, Oct 10 2018 *)

CurrentNum=10
PreviousNum=1
NewNum=0
def NextPalindrome(StartNum):
    FoundNext=0
    t=0
    n=0
    Number=0
    Lastdigit=0
    while FoundNext<=PreviousNum:
        n=n+1
        t=StartNum+n
        Number=t
        Reverse=0
        while Number>0:
            Lastdigit=Number%10
            Reverse=(Reverse*10)+Lastdigit
            Number=Number//10
        if t==Reverse:
            FoundNext=n
    return n
print(1)
print(10)
for x in range(50):
    NewNum=NextPalindrome(CurrentNum)
    print(NewNum)
    PreviousNum=CurrentNum
    CurrentNum=NewNum

A319795 a(n) = n^(n+1)/(n-1)^n for n>1, rounded to nearest integer.

Original entry on oeis.org

8, 10, 13, 15, 18, 21, 23, 26, 29, 31, 34, 37, 40, 42, 45, 48, 50, 53, 56, 59, 61, 64, 67, 69, 72, 75, 78, 80, 83, 86, 88, 91, 94, 97, 99, 102, 105, 107, 110, 113, 116, 118, 121, 124, 126, 129, 132, 135, 137, 140, 143, 145, 148, 151, 154, 156, 159, 162, 164

Offset: 2

Jim Singh, Sep 28 2018

Cf. A084363, A022852, A121384.

seq(round(n^(n+1)/(n-1)^n),n=1..100); # Muniru A Asiru, Sep 28 2018

for(n=2,30,print1(round(n^(n+1)/(n-1)^n),","))

A319727 Rounded frequencies of notes in the shruti scale of Indian classical music, starting with 260.7 Hertz for C-equivalent note.

Original entry on oeis.org

261, 275, 278, 290, 293, 309, 313, 326, 330, 348, 352, 367, 371, 391, 412, 417, 435, 440, 464, 469, 489, 495, 521, 549, 556, 579, 587, 618, 626, 652, 660, 695, 704, 733, 743, 782, 824, 834, 869, 880, 927, 939, 978, 990

Offset: 1

Jim Singh, Sep 26 2018

nonn

A shruti can be interpreted as the smallest interval of pitch the ear can detect and a singer or musical instrument can produce, and accordingly the 'Grama' system divides an octave into 22 parts.
The scale involves 256/243, 25/24 and 81/80 as fractions.
Note that ((81/80)^10) * ((256/243)^7) * ((25/24)^5) = 2.
The frequencies correspond to the ratios [1/1, 256/243, 16/15, 10/9, 9/8, 32/27, 6/5, 5/4, 81/64, 4/3, 27/20, 45/32, 729/512, 3/2, 128/81, 8/5, 5/3, 27/16, 16/9, 9/5, 15/8, 243/128, 2/1].
The start is A-equivalent note = 440 Hz. The initial term (rounded frequency of C-equivalent note) is calculated as (16/27) * 440 Hz = 260.7 Hz.

Wikipedia, Shruti (music)

Cf. A131071, A131062.

Ratios={[1/1, 256/243, 16/15, 10/9, 9/8, 32/27, 6/5, 5/4, 81/64, 4/3, 27/20, 45/32, 729/512, 3/2, 128/81, 8/5, 5/3, 27/16, 16/9, 9/5, 15/8, 243/128];}
a(n)={n--; round(440*16/27*2^(n\22)*Ratios[n%22+1])} \\ Andrew Howroyd, Sep 27 2018

A316742 Stepping through the Mersenne sequence (A000225) one step back, two steps forward.

Original entry on oeis.org

1, 0, 3, 1, 7, 3, 15, 7, 31, 15, 63, 31, 127, 63, 255, 127, 511, 255, 1023, 511, 2047, 1023, 4095, 2047, 8191, 4095, 16383, 8191, 32767, 16383, 65535, 32767, 131071, 65535, 262143, 131071, 524287, 262143, 1048575, 524287, 2097151, 1048575, 4194303, 2097151, 8388607, 4194303

Offset: 0

Jim Singh, Jul 12 2018

			Let 1. The first four terms are 1, (1-1)/2 = 0, 2*1+1 = 3, 1.
Let 4*1+3 = 7. The next four terms are 7, (7-1)/2 = 3, 2*7+1 = 15, 7.
Let 4*7+3 = 31. The next four terms are 31, (31-1)/2 = 15, 2*31+1 = 63, 31; etc.

Jim Singh and others, Fascinating periodic sequence pairs, Mersenne Forum thread, July 2018.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2).

Cf. A000225, A024036, A083420.

a:=[1,0,3];; for n in [4..45] do a[n]:=a[n-1]+2*a[n-2]-2*a[n-3]; od; a; # Muniru A Asiru, Jul 14 2018

seq(coeff(series((1-x+x^2)/((1-x)*(1-2*x^2)), x,n+1),x,n),n=0..45); # Muniru A Asiru, Jul 14 2018

CoefficientList[Series[(1 - x + x^2)/((1 - x) (1 - 2 x^2)), {x, 0, 42}], x] (* Michael De Vlieger, Jul 13 2018 *)
LinearRecurrence[{1, 2, -2}, {1, 0, 3}, 46] (* Robert G. Wilson v, Jul 21 2018 *)

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) for n>2, a(0)=1, a(1)=0, a(2)=3.
From Bruno Berselli, Jul 12 2018: (Start)
G.f.: (1 - x + x^2)/((1 - x)*(1 - 2*x^2)).
a(n) = 2*a(n-2) + 1 for n>1, a(0)=1, a(1)=0.
a(n) = (1 + (-1)^n)*(2^(n/2) - 2^((n-3)/2)) + 2^((n-1)/2) - 1.
Therefore: a(4*k) = 2*4^k - 1, a(4*k+1) = 4^k - 1, a(4*k+2) = 4^(k+1) - 1, a(4*k+3) = 2*4^k - 1. (End)

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User: Jim Singh

A320277 A digitized pure tuning tone, sampled at standard settings for consumer audio: a(n) = floor(sin(2*Pi*(440/44100)*n)*32767).

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A320170 Every pair of consecutive numbers sums to a palindrome, starting with a(0)=1; a(1)=10, and taking a(n) = smallest number > a(n-2).

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A319795 a(n) = n^(n+1)/(n-1)^n for n>1, rounded to nearest integer.

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A319727 Rounded frequencies of notes in the shruti scale of Indian classical music, starting with 260.7 Hertz for C-equivalent note.

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Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

A316742 Stepping through the Mersenne sequence (A000225) one step back, two steps forward.

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A320277 A digitized pure tuning tone, sampled at standard settings for consumer audio: a(n) = floor(sin(2Pi(440/44100)n)32767).